An Introduction to Mechanics of Materials
© Vijay Gupta
Lovely Professional University, Punjab
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Table of Contents
Summary 54
4 FORCES AND MOMENTS IN BEAMS 56 4.1 Introduction 56 4.2 Sign convention 57 4.3 Loads and supports 58 4.4 Determining shear forces and bending moments 59 4.5 General procedure for drawing shear force and bending moment diagrams by method of sections 61 4.6 The area method of drawing the SFDs and BMDs 64 Summary 69
1 STRUCTURES, LOADS AND STRESSES 4 1.1 Mechanics of material 4 1.2 Deformation and resisting forces 4 1.3 Other loadings, stresses and strains 5 1.4 The concept of stress at a point 7 1.5 Stress on oblique planes 11 1.6 Notation for stress: double-index notation 12 1.7 Equivalence of shear stresses on complementary planes 13 1.8 Stresses in a thin circular pressure vessel 14 1.9 Summary 15
5 STRESSES IN BEAMS 71 5.1 Introduction 71 5.2 Relating curvature of the beam to the bending moment 72 5.3 Composite beams 78 5.4 Stresses in beams carrying shear forces 82 5.5 Relating shear stresses to the shear force in a beam 83 5.6 Shear flow in beams 86 5.7 Shear centre 88 5.8 Plastic deformations in beams 88 5.9 Strain energy in bending 89 Summary 90
2 DEFORMATIONS, STRAINS AND MATERIAL PROPERTIES 17 2.1 Fundamental strategy of mechanics of deformable mechanics 17 2.2 Statically indeterminate problems 20 2.3 Lateral strain: Poisson ratio 23 2.4 Shear strain 25 2.5 Thermal Strains 27 2.6 Tensile test 28 2.7 Idealized stress-strain curves 30 2.8 Pre-stressing 32 2.9 Strain energy in an axially loaded members 33 2.10 Calculating deflections by energy methods: Castigliano theorem 33 2.11 Strain energy in an elastic body 38 Summary 39
APPENDIX B PROPERTIES OF AREAS 93 B.1 First moments of area and centroid 93 B.2 Second moments of area 93 B.3 Parallel axes theorem 94 B.4 Perpendicular axes theorem 96
3 TORSION OF CIRCULAR SHAFTS 41 3.1 Introduction 41 3.2 Relating angle of twist to twisting moment 41 3.3 Stresses and strain in a circular shaft 43 3.4 Hollow shaft 46 3.5 Statically indeterminate shafts 47 3.6 Composite shaft 49 3.7 Torsion of thin-walled tubes 50 3.8 Plastic deformation in torsion 51 3.9 Limit Torque 52 3.10 Strain energy in torsion 53
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3
Clearly, this part of the beam is not in equilibrium with just the external force, P. We need additional external (i.e., external to this part of the beam) forces and/or moments. The open arrows in Fig. 1.1b 1.1b show the external force and moment moment required to balance the applied load P. We will for the time being refer to these as the reaction forces and the moments. Where do these forces and moment come from? As we apply the force P to the beam and if these reactions do not kick in, the beam will tend to shear from the stump built into the wall at the leftend. The distortion of the beam so produced results in generation of material forces within the beam that resist this shearing action. When we consider the part of the beam shown in the free body of Fig. 1.1b, these material forces appear as external forces (and moments) on the beam. beam. Of course, there are equal equal and opposite reaction on the the stump of the beam built in the wall.
1 Structures, loads and stresses
1.1 Mechanics of material The subject matter of a course on mechanics of materials deals with structures. A table or a chair is a structure. A building building is a structure. A bridge is a structure. A TV tower is a structure. So is a printed circuit board, the casing of a fax machine, or the body of a car. Among the many purposes of the various structures, structures, one common purpose is to resist and/or and/or transmit forces acting on it. By resisting a force we mean that the structure would not break under the force. The structure of a building is designed to resist the loads which include the weight of the people and things occupying it, the forces of wind acting on it in a storm, even the load imposed by an earthquake, and the self-load of the building itself. itself. The structure of an aeroplane resists the aerodynamic loads, the weight of its occupants (including the dynamic loads during acceleration and deceleration), the load imposed by the thrust produced by the engines, and of course the weight of the structure itself.
We can summarize the above as:
The external forces acting on a structure result in deformation of the structural members. The deformation so caused result in resisting forces within the material of the members. When we consider the equilibrium of a part of the member, these internal forces come into play as external forces and balance the applied forces or moments.
1.2 Deformation and resisting forces
How does a structure resist loads?
Consider a vertical rod anchored as shown in Fig. 1.2. It is common knowledge that when you apply a longitudinal force P to this rod, it elongated a definite amount (depending on its dimensions and its material). Consider a portion of the rod enclosed by the broken line rectangle. The free body diagram (FBD) of this part is shown in Fig. 1.1b. Since the rod is in equilibrium after the elongation, there must be a force that balances the applied force P. Where does that force come from? Clearly, there are internal internal forces which are holding this part from running away from the rest of the member. These internal forces as seen in the previous section are the consequence of the distortion produced in the bar. Now
Consider a simple case of a cantilever beam loaded loaded as shown in Fig. 1.1. If the beam is in equilibrium, the net force or moment on the beam or on any part of it must be zero. Let us consider the part part of the beam within the dashed-line box shown. This is known as a free-body diagram (FBD).
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more the force we apply, more is the elongation, suggesting that the resisting force that develops in the rod depends on the elongation. Robert Hooke, a British scientist is credited to be the first to explore the relationship between the resisting force force and the elongation. elongation. He found out in 1678 that for a given material, the resisting force does not depend on the elongation but on the relative elongation produced. He introduced the term strain to denote the elongation relative to the original length of the bar. If l denotes the original length, and δ the elongation, the strain is defined by Strain,
materials. Steel has about the largest value of the the elastic modulus of about 200 1 GPa . Cast Iron has about about half this value. Aluminium is still lower lower at 70 GPa. The summary statements of the previous section can now be recast as:
The external forces acting on a structure result in strains i n the structural members. The strains so produced result in stresses within the material of the members. The stresses, for the most part, are proportional to the strains produced. The constant of proportionality is termed as the modulus of elasticity.
1.3 Other loadings, stresses and strains Fig. 1.2 A bar loaded longitudinally
(1.1)
Strain is dimensionless and has no units. Hooke also found out that it is not the force, but the intensity of force measured as the force per unit area that needs to be considered. considered. He called it stress. If A is the area of cross-section of the bar, the stress is defined as the resisting force P divided by A. Stress,
Fig 1.3. The columns supporting a highway deck, the boom of a crane and a foundation block are all structural members in compression
(1.2)
Stress has the dimensions of force per unit area (hydraulic pressure, too, has the same dimensions) and has SI units of Newton per meter squared (M/m2) which is termed as Pascal and abbreviated as Pa.
We had in the previous section considered one kind of load that tends to elongate a member, leading to one type of strain (longitudinal) and one type of stress (tensile). It is possible possible to load members members in various various other ways. ways. Compression load is a familiar example. Compression results when two bodies are pressed together. Columns that support elevated highways (Fig. 1.3), water tanks or roofs are all compression members. A compression member shrinks in length because because of the load, resulting in compressive strains and stresses. The footing of a machine is also under a compressive load. So also is the boom of the crane.
He further found that the stress and strain, in a large part, have a simple linear relation for bars made of the same material: Stress
Strain, or,
(1.3)
The constant of proportionately, E , is termed as the elastic modulus, and depends on the material of the bar. Strain being dimensionless, dimensionless, the dimensions (and E are the same as those of stress. units) of E
Compression also results in in the situation shown in Fig. 1.4. Here two plates are riveted together. As a force is applied to the plates that tends to to pull them apart, the rivet compresses the plates at the rivet holes as shown. The compressive stresses so produced in the plates are also termed as the bearing stresses.
Combining Eqs. 1.1 -1.3, we get:
(1.4)
The value of the elastic modulus E for most construction materials is quite high, denoting that it takes fairly large forces to produce small elongations. Table B.1 in Appendix B gives the values of the elastic modulus for some common
1
5
A GPa is 109 Pa, or 109 N/m2
Another type of load is the shear load. Consider a block of rubber glued to a table on one side and a board on the the other (Fig. 1.5). If we apply a load P as shown, the block of rubber will undergo distortion and it will tend to slide off the table. The distortion results in what what are termed as shear strains , which, in turn, result in shear stresses in the material.
compressive stresses therein, while the lower fibres elongate introducing tensile strain and tensile stresses therein. The magnitude of these compressive and tensile stresses is such that they integrate out to zero, which they must, since there is no applied force on the bar.
Shear stress and shear strain is also produced in the rivet of Fig. 1.4. As the force is applied the rivet has a tendency to shear at the middle. Shear stresses develop to counter this action and keep each half the rivet in equilibrium.
Bearing stresses in the plate
Shear stresses in the rivet Fig 1.6 Twisting of a shaft. The shear stresses in the cross-section of the shaft give rise to a resisting twisting moment.
Fig. 1.4 Ri veted plates plates
But the resultant moment is non-zero as is required to resist the applied bending moment and ensuring equilibrium. Shear strain and shear stresses are also produced when a shaft is twisted as in Fig 1.6. If we consider the shaft as an assembly of thin slices stacked together, the twisting action of the shaft tends to make the slices slip on one another. Internal
Let us summarize what was learnt in this section:
P
Forces that tend to reduce the size of a structural member produce compressive strains which, in turn, produce compressive stresses. Within limits, the magnitude of compressive stresses vary linearly with the compressive strain, the constant of proportionality being the same as the modulus of elasticity, E introduced in the last section with tensile stresses and strains.
Fig 1.5 Shear in a block of rubber
forces develop which resist these motions. motions. These internal resisting forces are the shear stresses and the resultant of these is a moment, termed as the twisting moment .
Fig. 1.7 Bending of a beam.
Another type of distortion occurs when a moment is applied to a bar which tends to bend it. As shown in Fig. 1.7, the upper fibres of the beam tend to shorten from their original length, introducing compressive strain and resulting in
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uniformly on this material. Consider an elemental area 2 δA on a face of a structure exposed by making making a cut at that location. location. Let δF represent the force that the material that has been removed was applying at the elemental area δA. This force arises from the deformation of the material under whatever external load is being applied to the structure.
Forces that tend to distort the shape of a member, as in Fig. 1. 5, produce shear strains which in turn produce shear shear stresses. The more the shear strain, the more is the shear stress. A twisting moment applied to a shaft produces shear strains in the shaft. These shear strains give rise to shear stresses across the cross-section of the shaft, which result in a moment which balances the external twisting moment. The more the external moment, more is the strain, more the stress, and more the resisting moment. A moment tending to bend a member as in Fig. 1.7, produces both compressive and tensile strains and stresses stresses in the beam. These stresses give rise to a resisting moment which balances the bending moment.
The stress at a point is defined as the intensity of the internal force of deformation at this point: Stress vector on this face at this point is t
The stress vector t depends upon the location as well as the direction of the surface. If we had made the cut (at the same point) with a different inclination (i.e., with the outward normal in some other direction), the stress vector would have been different as is shown in Fig. 1.9
1.4 The concept of stress at a point In Section 1.2 we were dealing with a very simplistic case of a straight bar of area A with a longitudinal load P and we defined stress simply as load per unit area. This assumes a uniform distribution of stresses – a very severe assumption. In practice, the deformations deformations due to load and, consequently, consequently, the strains and stresses will vary from point point to point. It is, therefore, convenient to define stress as the intensity of force at a point.
(a) A solid object
(b) Section through X through X along along bb (c) Section through X through X along along cc Fig. 1.9 Stress depends on the direction of the cut
Area δ A
We can take the component of this vector along the normal to the surface and along the surface: δ F
F
The component of the stress vector normal to the surface is termed as the normal stress and is denoted by the Greek letter ζ. The component of the stress vector along the surface is termed as the shear stress and is denoted by the Greek -2 letter η. As noted before, the dimensions of stress are Force/Area, or FL , or ML 1 -2 2 T , and its SI unit is N/m or Pascal, Pa. A Pascal is a very small small unit and it is common to have stresses in kPa or MPa.
F
(a) A gear
(b) A cut-out portion of the tooth
Fig. 1.10a shows a bar loaded uniformly in tension as shown. Because of this load the bar elongates setting up axial strains and stresses along any cross section. Fig. 1.10b shows the resulting free body. It stands to reason that that if we take a section of this bar at any level, the resulting distribution of stresses must be the same. However, if the l oading is not uniform but concentrated at a point in
Fig. 1.8 Stress at a point
Consider a gear which is meshing with another gear which applies a force P on a tooth (Fig. 1.8). Let us consider the equilibrium of this tooth. Fig 1.8b shows the part of the tooth as a separate separate free body. The force that balances the external force F is the force which was binding this tooth to the rest of the gear. gear. This is an external force that arises because of the loading of the gear and its consequent straining. Let us imagine this straining straining and the internal internal forces are not distributed distributed
2
Area is a vector quantity, its direction denoted by the direction of the outward normal to the surface.
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the bar as shown in Fig. 1.11a the story changes. There is no reason to expect that the stress distribution at a section is uniform. In fact, it is not. Moreover, it changes as we move from section to section. If we take take a section section at different different heights along planes b, c, or d, the resulting free bodies are shown in Fig. 1.11b, c and d. There total of the internal (a) (b) forces or stresses at each of these sections Fig. 1.10 The stress distribution must be equal to P to balance the applied on a uniformly loaded bar force. But the distribution of these stresses is quite different at the three sections. The distribution of stresses becomes more uniform as we move upwards, away from the point of application of the force P.
distributed loading) is uniform everywhere, and therefore, far away from the point of concentrated loading in Fig. 1.11, we can take the stresses to be uniformly distributed. We shall, in this elementary text, will routinely make the assumption of uniform stresses across sections, unless the context of the problem forbids it.
Example 1.1 Bearing stresses in foundations Consider a wooden column (Fig. 1.12) resting on a concrete footing. Determine the maximum value of the load P if the maximum permissible stress in concrete is 60 MPa and in wood is 25 M Pa. Solution: We assume that the stresses at any cross-section are distributed uniformly. If P is the external load , the stress in the wooden column is P (N)/ [π× (0.010 m) 2 ] = 3,183 P (N/m2 or Pa). Similarly, the stress in 2 the concrete footing is P (N)/ [(0.030 m) ] = 1,111 2 P (N/m or Pa). Equating these stresses to 25 MPa and 60 MPa, respectively, we find the value of P from the first as 7.85 kN, and from the second as 54 kN.
In fact, there is a principle known as St. Venant principle , named after a nineteenth century French theoretician, which lays down this behaviour: The difference between the stresses caused by statically equivalent load systems is insignificant at distances greater than the largest dimension of the area over which the loads are acting.
Clearly, the maximum permissible load will be the lesser of the two, two, i.e., 7.85 N. This suggests that we can reduce the footing size drastically, if bearing this load is the only design consideration.
d
Fig. 1.12 Further we have considered only the compressive stresses. Thin columns may buckle under compressive load much earlier earlier than they fail in compression. We will need to check that too. We shall deal with this in a later chapter.
c b
P (a)
P
P (b)
(c)
P (d)
Fig. 1.11 The stress distribution becomes uniform as we move away from the point of loading
The loading in Figs. 1.10 and 1.11 are statically equivalent since they result in the same net force and moments on the structure. structure. The stresses in Fig. Fig. 1.10 (with
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∑ ∑
Example 1.2 A truss A truss structure is a load bearing structure assembled such that its members carry only axial loads, either tension or compression. It is made up of straight, two-force members 3 joined together by frictionless pins. All the loads are applied applied at only the joints. Further, there are no external external moments applied to such structure.
(b)
Solving the two simultaneously, we get TBA = 0.51P, and TBC = 0.78P The stresses in the bars are obtained by dividing the above load by the cross-sectional area (assuming uniform stress distribution), which is π× (0.003 m) 2 = 2.83×10-5 m2. The stress along BC will be the limiting stress, since it is higher. Thus,
A truss is a gross approximation approximation to an actual structure. There are no practical frictionless pins, and most members members carry some bending loads as as well. But the advantage of considering only two-force members (in simplifying calculations) is so enormous that we make the truss approximation wherever feasible.
Fig. 1.13 shows a pinned structure consisting of two bars AB and BC from which a weight P hangs P hangs as shown. If the diameter of each each bar is 6 mm and the permissible stress in the bar is 80 MPa, find the value of the maximum load P
This gives the maximum load P as 2.9 kN
A
1m
B C 20kN 1m
Fig. 1.14
Example 1.3 A truss with roller support
. .
Consider a structure structure shown in Fig. 1.14. If the members of the structure are mild steel with a radius of 20 mm, find the stresses in members AB, BC and BC and AC .
Solution: Consider the FBD of point B where the load P is being applied. Since AB is a pinned member, and hence a twoFig. 1.13 force member there can only be tension along it. Let us call it T BA. Similarly, there will be tension T BC along the member BC.
Solution: The support at point A is termed as a hinged support . This means that the support cannot apply any moment at this point to the structure and the structure can articulate at this point. The only possible external reactions reactions (from the support) at this point is a force with components both along and normal to the surface. There is no reaction moment at this point. The support at point B is termed as a roller support. A roller support, besides permitting rotation of the structure, permits translation of structure along the supporting surface. surface. This means that not only is the reaction moment zero, but the force component along the support surface is also zero. The only reaction possible at such a support is a force component normal to the surface.
The equilibrium of a body requires that the vector sum of forces and moments should vanish:
∑ ∑
(a)
(1.5)
We need to apply only the force equations equations here. Taking the component-wise sum of the forces, we get
3
A member on which external forces act only at two distinct points (and there is no external torque acting on it) is termed as a two-force member. The forces acting on a two-force member are collinear and opposed .
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Fig. 1.15b shows the forces on the member AC . It is clearly that the the net force on 4 the member is along the length of the member, tensile, and equal in magnitude
To find the stresses in the members, we need to determine the forces on them by
√
20 kN
R Ay
to of 20 kN.
A
20 kN
R Ax
Further, since at point B there is no vertical force, there cannot be any 5 force in member AB.
C
R Bx
20kN 20kN
C
The stress in the two members can be found by dividing the forces with the respective crosssectional areas. Thus, (a) (b) (c) Fig. 1.16 Shear stresses in a key the stress in member AC is 28.3 (kN)/(π/ (kN)/(π/ 4)(15×10 4)(15×10 3 m)2 = 160.1 MPa, tensile, and that in member BC it is − 20 (kN)/(π/ (kN)/(π/ 4)(15×10 4)(15×10- 3 2 m) = 113.2 MPa, compressive.
20kN 1m
(b) Force along member AC member AC
(a) FBD whole structure Fig. 1.15
doing a statical analysis. The first step in solving this problem consists of determining the reactions at the supports A and B. The reactions at support are found by considering the equilibrium of the FBD of the whole structure structure as drawn in Fig. 1.15a. Here all the external forces have been shown: the applied load of 20 kN , the horizontal and vertical reactions R Ax and R Ay at the pinned support A, and the horizontal reaction R Bx at the roller support B. The structure should be in equilibrium under the action of these forces. There are two conditions for equilibrium: vector sum of all the forces as well as all the moments must be zero.
Example 1.4 Key Consider the transmission of power by belt and pulleys. Fig. 1.16 shows a pulley being driven clockwise by a belt. If the pulley is turning at 100 RPM and and if the power transmitted is 1 kW, what is the shear stress in the key? The shaft dia. is 20 mm, and the dimensions of the key are 4 mm × 4 mm and a length of 25 mm.
Writing the sum of horizontal and vertical forces as zero gives us:
∑ ∑ N
Solution:
(a)
A key is a common device used to couple a pulley with a shaft, so that as the pulley rotates the shaft rotates with it. It is a metal piece inserted so that a part of it is inside the shaft (in a slot termed as a keyway), and a part is within the pulley. As the pulley turns, the key moves with it. The pulley applies a force on the key towards the right as shown in Fig. Fig. 1.16b. The key whose lower part is enmeshed enmeshed with the slot within the shaft makes applies a force on the shaft to make it turn with the pulley. As the key pushes the shaft, the shaft, in turn, pushes the key back towards the left.
(b)
Eq. (b) gives R Ay = 20 kN. There is only one equation equation to determine the other two two unknowns. We need one more equation. equation. That we get from the moment moment balance. We can take the moments of forces about any convenient point. Here we take it about the point A (because then the two force components at this point contribute no moment):
∑
The member BC is under a compression
(c) 4
This should have been obvious without the calculations since members AC and BC are two-force members. The forces acting on a two-force member are collinear and opposed. 5 This is quite interesting. interesting. If there is no force in member AB, we really do not need that member. member. Can the structure structure survive without without the member AB? The answer is: yes, if we are concerned only with the equilibrium, and no, if we worry about the stability.
This gives = 20 kN, and, then Eq. (a) gives 20 kN, that is, 20 kN in a direction opposite to that shown on the FBD in Fig. 1.15a, or 20 kN towards left.
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Now, if we look at Fig. 1.16b, we notice that the upper part of the key is being pushed to the right while the lower part of the key is being pushed to the left. We may say that the key will have a tendency to shear in the middle. middle. Fig. 1.16c shows the FBDs of the lower and the upper parts parts of the key. The internal shear stresses acting on each half balance the externally applied forces.
Further, the pull force will tend to shear the plates as shown in Fig. Fig. 1.16d. This -5 2 shearing action acts on a total area of 2×(10 mm × 2 mm) = 4×10 m for the force in each rivet, which which is 1 kN. Thus, the shear stress in the the plates will be 1 -5 2 kN/4×10 m = 25 MPa. Further the pull force will tend to crush the area of the plate as shown in Fig 1.16e, causing compressive or bearing stresses in the area in immediate contact of the rivet. Surely these will not not be distributed uniformly over the area. area. But if we make the assumption that they are, the resulting bearing stresses will be 1 kN/(2 mm × 10 mm) = 50 MPa.
Since the power transmitted is I kW, and the RPM is 100, the torque is 1000 W/ the key (at (2π rd/rev×100 rev/min/ 60 s/min) = 95.5 9 5.5 N.m. The force acting on the a radius of 10 mm) which transmits this torque is obtained as 95.5 Nm/0.01 m = 9,550 N The total shear force acting on the key (the lower or the upper part) is, thus, 9,550 N. The area on which it acts is 4 mm × 25 mm and, therefore, the shear stress is 9,550 N/(4×10-3 m × -3 25×10 m) = 95.5 MPa.
1.5 Stress on oblique planes Consider again a bar of area A loaded axially as shown shown in Fig. 1.18a. Let us look at the stresses on an oblique plane b inclined at an angle6 . It is clear that if we draw the FBD of the lower portion of this bar as in (b), the total internal force acting on this section is P as shown. But what is the stress here? The area on which this force acts in not A but larger than A, equal to A /cos /cos . If we assume, as before, that the stress is distributed uniformly over the area (which shall be quite true if the plane is not to close to the point of application of the load) the stress vector here will be P divided by , or t = Pcos/A. cos/A. We can resolve this stress vector in two components (Fig. 1.18c), one normal to the oblique area, and the other along it. The normal component is the tensile stress: stress:
As an aside, note that the applied load is producing compressive or bearing stresses as well. The area on which the compressive force (9,550 Fig. 1.17 Lap joint N) acts is 2 mm × 25 mm and, therefore, the bearing stress is 9,550 N/(2×10-3 m × 25×10-3 m) = 190.1 MPa.
(1.6)
This is also termed as the normal stress. The tangential component is a shear
Example 1.5 Riveted lap joint Consider two steel plates 2 mm thick joined together as shown in Fig. 1.16(a) and (b) by two rivets of 10 mm diameter. The centres of rivets are 10 mm from the edge of the plates. Determine the stresses in the rivets if a force of 2 kN is applied to the plates trying to pull them apart.
P η
θ
ζ
-η
ζ
-η
η
b
Solution:
P
As was discussed earlier, the pull force tends to shear the rivets as shown in Fig. 1.17c. Since there are two rivets, we can assume at each rivet sustains half the total force, i.e., 1 kN. This is the shear force in each rivet acting on the crosssectionl re of π× (0.005 m) 2 = 7.85×10-5 m2. Therefore, the the shear stress in rivet shanks is 1 kN/7.85×10-5 m2 = 12.7 kPa.
(a)
P (b)
(c)
(d)
(e)
Fig. 1.18 1.18 Stresses on an oblique plane
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The ngle θ is lso the ngle between the normls to the new plne nd the original plane.
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stress:
.
width of the strip as we go around the tube through one circumference, 2πR 2πR as is, therefore, shown in Fig. 1.19b. This angle is,
(1.7)
Clearly, the value of the tensile and shear stresses at a location in the bar varies with the angle of inclination of the plane being considered. The maximum tensile stress occurs when is is zero, i.e., the cut is perpendicular to the axis of the increases. But the behaviour of shear stress is bar, and reduces continuously as increases. sin quite interesting. interesting. If we note note that cos cos sin is (1/2)sin2 (1/2)sin2 , we see that the shear stress η first increases as increases, attains a maximum value (equal to P /2 A, i.e., half the maximum value of the tensile stress) when is π /4, and then decreases back to zero as increases to π /4. Also note that if the sectioning plane is inclined in the other sense (with negative values of of ), ), the shear stress on it has a reversed sense too. This is also obtained from the expression expression for η given above since now now is negative.
.
The compressive stress in the cardboard tube along the axial direction is 100 N/ [2π [2π (35×10 (35×10-3 m)2×(2×10-3 m)] = 227 kPa. We can now use the Eq. 1.7 to determine the shear stress along the seam, which is a surface with the normal in the direction as shown in the figure. It can be seen that the angle here here is the same as the angle in Eq. 1.7 (the angle between the normals to the axial plane and the new plane on which stresses are to be determined). Therefore,
̂
= 57.7 kPa
Two planes, one at angle , and the other at angle - are termed as complementary planes. Shear stress on complementary planes have same magnitude but opposite sense .
This is the shear stress stress along the glued seam. seam. The seam would rip if the the glue cannot sustain this level of shear stress.
Please note that the equations obtained above are valid only for the special case of axial loading of a straight uniform bar. If the geometry or loading loading were different, these would no longer be valid.
1.6 Notation for stress: double-index notation
Example 1.6 A Cardboard tube
As is clear from the discussion above, the stress vector at a point depends on the orientation of the surface under consideration. We can, at the same point, consider many differently inclined surfaces. The Fig. 1.20 Specification of stress components orientation of a surface is denoted by the direction of the outward normal to the surface. Therefore, we must specify the direction of the normal to the surface whenever we talk of stress at a point.
A cardboard tube of diameter 70 mm is made from a strip of width 60 mm and thickness 2 mm wound spirally with the edges glued together as shown in Fig. 1.19a. The cylinder is subjected to an axial load of 100 N, determine the shear stress in the glued joint.
Solution: The angle of the spiral can be obtained by imagining that as the cardboard strip is wound up, the strip should advance in the axial direction a distance equal to the
We shall see later (in Chapter 6) that if we know the stress vectors on three mutually perpendicular planes we can determine the stress vector on any other plane. We, therefore, need to specify stresses stresses on three such planes. planes. Figure 1.20 shows stresses on three mutually perpendicular planes, with normals in x, y, and z directions. We have shown here, the components of stress vectors vectors on each of these three planes. The tensile (or the compressive) components are represented by the symbol ζ, and shear stress components by the symbol η. We use here a double-index notation, with the first index of a stress component denoting the plane on which it acts, and the second index denoting the coordinate direction of
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a component itself. Thus, η xy is a component of shear stress acting on a plane with the normal in the x coordinate direction, the component itself being in the y coordinate direction. Similarly, η zy is a shear stress component acting on a plane with the normal in the z coordinate direction, the component being in the y coordinate direction, and ζ zz is a tensile stress component in the z direction acting on a plane with normal in the z direction. It should be apparent apparent that there are:
It may be verified that all the stress components shown in Fig. 1.20 are positive.
Example 1.7 Nomenclature and signs of stresses Fig. 1.21 shows stress components on some faces labelled (a) to (d ). ). Name these planes on which the stresses have been shown. Name the stresses and determine their signs according to the sign convention outlined above.
three tensile components, ζ xx , ζ yy , and ζ zz, one each on the three faces (note that the indices are repeated in each of them), six shear stress components: two on each face, η xy and η xz on the x-face, η yx and η yz on the y-face, and η zx and η zy on the z-face. We shall soon see that three of them are equal to the other three, so that there are only three independent shear stresses at a point (see section 1.7).
Sign convention: It is convenient to use the following sign convention:
Face c
3
F a c e d
2
6 4 5
a c e a Face b F
1
10
y
11 12
x Fig. 1.21
Solution:
Plane a is an x-plane, b a y-plane, c a z-plane and d is an x-plane (with the normal in – ve ve x direction. The nomenclature and signs of the stresses are tabulated below8: Stress
A stress component is considered positive when a positively directed force acts on a positive face, or a negatively directed force acts on a negative face 7 .
1 2 3 4 5 6 7 8 9 10 11 12
In other words, if both the sign of the force and the face are the same, positive or negative, the resulting stress is a positive and if the two signs differ, the stress is negative. This is summarised in Table 1.1 below: Table 1.1 Sign convention for stress Direction of force In positive coordinate direction(+) In positive coordinate direction(+) In negative coordinate direction(−) In negative coordinate direction(−)
9
7
z
We first define a face to be positive or negative: a face with the outward normal in the direction of the positive coordinate axis is termed as a positive face; else it is termed as a negative face. When a positively directed force acts on a positive face or a negatively directed force acts on a negative face, the stress is assigned a positive sign. And when a negatively directed force acts on a positive face or a positively directed force acts on a negative face, t he stress component is assigned a negative sign.
Direction of normal In positive coordinate direction (+) In negative coordinate direction (−) In positive coordinate direction(+) In negative coordinate direction(−)
8
Sign of stress Positive(+)
Positive(+)
Index of plane x x x z z z y y y x x x
Index for force component x z y x z y z y x x z y
Symbol for stress ζ xx η xz η xy η zx ζ zz η zy η yz η xz ζ zz ζ xx η xz η xy
Sign of plane + ve + ve + ve + ve + ve + ve + ve + ve + ve − ve − ve − ve
Sign of force component − ve − ve + ve − ve − ve + ve − ve + ve − ve + ve − ve − ve
Sign of stress − ve − ve + ve − ve − ve + ve − ve + ve − ve − ve − ve + ve
Negtive(−)
1.7 Equivalence of shear stresses on complementary planes
Negtive(−)
7
8
Note that the signs of stresses on opposite faces a and d are identical, as they should be. In fact, the sign convention has been designed to ensure this.
This sign convention follows from the consideration of action and reaction having the same sign.
13
Consider a small two-dimensional element of dimensions δx and δy and of unit depth as shown in Fig.1.22. We have ζ yy drawn the stress components on the four faces of the element. element. We assume η yx η xy a two-dimensional state of stress, i.e., δ y we assume that there is no loading and, ζ xx ζ xx hence, no stress component in the third δ x η xy direction. We shall now consider the y η yx equilibrium of the element under the ζ yy x action of the forces due to these stresses:
•
1.8 Stresses in a thin circular pressure vessel 9
Consider a thin cylindrical pressure vessel of length L, radius R, and wall L thickness t . Fig. 1.23 shows the cylinder with with end plate removed. removed. A cylindrical polar co-ordinate system with r , , and z coordinates is ideal for this geometry. Let the internal excess θ R r t pressure be p. Under the action of z these forces, stresses will be set up in the cylindrical vessel. Following the development of the previous sections, there will be six stress components Fig 1.23 A thin cylindrical vessel with end plate removed required to describe the state of stress in the vessel walls: three tensile components: ζ rr , and ζ r , ζ z z , and ζ rr , ζ zz, and three shear stress components: ζ r zr .
Fig. 1.22 A 2-D in finitesimal element
Face Area (identified by the assume unit direction of depth outward normal) x −x y −y
Force x-component
δy•1 δy•1 δy•1 δy•1 δx•1 δx•1 δx•1 δx•1
ζ xx• δy −ζ xx• δy ζ yy• δx −ζ yy• δx
x-component
η xy• δy −η xy• δy η yx• δx −η yx• δx
When we consider the force balance, we verify that there is no net force on the x- and y- forces sum out to zero independently. We next take element: the sum of xthe sum of the moments moments about the centre point point of the element. Note that the tensile forces produce no no moment about that point. However, the shear forces do. The shear forces on opposite faces are equal in magnitude but opposite in sign and, therefore, constitute two couples, one anticlockwise (positive) and the other clockwise (negative). The moments of the two couples couples are found by multiplying the magnitudes of the forces and the perpendicular distance between the two lines of action. The moment balance equation gives: gives:
Let us first consider the tensile stress component ζ indicated by the first . As indicated subscript, , it acts on a plane with normal in the direction. This stress is exposed (or mde ‘externl’) ‘externl’ ) by taking a section as shown by a diametrical cutting plane as shown. Figure 1.24a shows the FBD of the lower half of the cylinder (ends removed). We can determine the stress component ζ by considering the equilibrium of the vertical component of all forces. The total vertical force due to the tensile stresses is ζ times the area on which L×t ). Lt . This force is these stresses act, which is 2×( L ). Therefore, this force is 2ζ 2ζ being balanced by the vertical component of the pressure forces acting on the inside the half-shell. The integration of the vertical component of pressure forces acting on this half of cylindrical shell is not straightforward. But it can be determined quite easily by resorting to a frequently used trick. The pressure force acting on the shell is equal and opposite to the pressure forces acting on the gas at the shell wall (the principle of action and reaction). Consider Consider the ‘FBD’ of the ‘gs’ contined in this hlf shell s shown in Fig. Fig. 1.24b. The verticl net pressure forces on the curved surface this FBD (which is the same and opposite to the
( )( )
,
which on simplification gives
=
three shear stress components η xy, η yx and η zx.
We shall show in Chapter 6 that we can, from these six components on three orthogonal planes through a point, determine the stress vector on any plane through that point.
.
This is an important result and establishes that the shear-stress components on adjacent (orthogonal) faces are equal in magnitude. We can, similarly, show that = and = . As much was stated in Section 1.6 without proof. proof. We reiterate what was stated there:
The state of stress at a point can be established with six components of stresses:
three tensile components, ζ xx , ζ yy , and orthogonal faces, and
ζ , zz
one each on the three
9
‘Thin’ refers to the condition tht the wll thic kness t is much less than the radius R.
14
pressure forces of Fig. 1.24a) is balanced by, and hence equal to the pressure force on the the diametrical plane of this FBD. This is p(2 RL). Therefore, the integrated pressure force on the curved surface of the FBD of Fig. 1.24a is 2 pRL.
class of arguments termed as the symmetry arguments. Consider the two halves of the cylinder as shown in Fig. Fig. 1.26. We have shown the shear stress components η r r on the upper half . Note that on the left-hand side, the plane has a normal in the +ve direction (being counter-clockwise), and the stress is in the +ve r direction. Therefore, the stress is η r r and is +ve. On the right-hand side of the upper half, both the signs are negative and therefore the stress is +ve again. The two stresses having opposite sense is correct, since the horizontal forces acting on this FBD should some out to zero (the sum of the horizontal component of uniform pressure forces can safely be assumed to be zero).
Lt = 2 pRL, or, 2 ζ
(1.8)
This tensile stress in the circumferential direction is also termed as the hoop stress.
Now on the lower half, the Newton third law (the principle of action and reaction) dictates that the stresses should have a sense opposite to that on the upper half. This is as shown. But here we run into a problem. The stresses are both outwards on the upper half, and both inwards on the the lower half. This violates symmetry. It is easy to see that it is not possible to distinguish between the upper and the lower halves of the cylinder. If somebody came and switched the two halves while a reader was away, there is no no way by which the reader can tell which is which. This symmetry requires that the state of stress on the two halves must be the same, either both inwards or both outwards. outwards. Thus, the requirements of third law and that of symmetry are in contradiction, and the both must hold! The resolution this contradiction is possible only if the stress η r r vanishes. This is a sufficient proof. We can construct similar symmetry arguments arguments to show that the other two shear components η rz rz and η z too must vanish. Thus,
We next consider the tensile stress ζ zz which is the stress on the plane with normal in the axial σ direction. Such a face is exposed when we cut the cylinder σ p with a r- plane parallel to the ends as shown in Fig. 1.25 which shows the FBD of one part of the vessel so (a) (b) exposed. The Fig 1.24 FBD of one half of thin cylinder exposing ζ external forces in the z-direction acting on this FBD are due to the tensile stress ζ zz acting on area which can be approximated by 2πRt 2πRt , and the pressure p acting on the end plate area of πR of πR2. The 2 z-force equilibrium, then, gives: ζ zz×2πRt - p× πR . This gives: θθ
θθ
η r r = η rz z = 0 rz = η
1.9 Summary
(1.9)
The third tensile stress is ζ rr radial direction. Note that on the inner rr in the radial curved surface the pressure p acts radially and, therefore, is the stress ζ rr rr at that point. But on ζ ZZ the outer curved surface, the pressure is zero, so the stress ζ rr rr is zero. Therefore, ζ rr rr varies from 0 to p across the thickness of the cylinder. In p any case, since r >> t , the value of ζ of ζ rr rr (between 0 and r ) is much less than ζ zz or ζ . Therefore, in comparison to the other tensile stresses in the cylindrical shell, it is common to neglect ζ rr rr ,
(1.10)
Let us next consider the shear stresses. We first discuss η r We shall use a very interesting interesting r .
(1.11)
Fig. 1.25 FBD for determining ζ zz
15
The external forces acting on a structure result in strains i n the structural members. The strains so produced result in stresses within the material of the members. The stresses, for the most part, are proportional to the strains produced. The constant of proportionality is termed as the modulus of elasticity which is a property of the material of which the structural member is made. In general, the stresses vary from point to point, but the use of the St. Venant principle permits us in many simple situations to assume uniform stress distribution across a section. This is a useful particularly at sections fairly distant from the points of application of loads. Stress vector at a point describes the intensity of internal forces that develop within a material in response to distortions that are produced under the application of external loads.
The stress vector at a point depends on the orientation of the surface on which the stress acts. Eqs. 1.6 and 1.7 describe the formula for calculating the tensile and shear stresses on an oblique plane in the case of longitudinal loading of a bar. The stress at a point can be described by stating the stress vectors on three mutually perpendicular planes with a total of three tensile components and six shear components. A double index notation is used to designate stress components, the first index representing the direction of the outward normal to the surface, and the second index the direction of the component itself. A stress component is considered positive when a positively directed force acts on a positive face, or a negatively directed force acts on a negative face. Equilibrium of forces require that the shear stress components occur in pairs, so that the shear stresses on complementary planes are equal: = , = and = . Thus, there are only three independent shear-stress components. We showed by symmetry arguments that the shear stress components in a thin walled cylindrical pressure vessels all vanish.
16
different slopes collapse into one line, as long as the material of the various bars , where the is the sme. This fct ws stted stted s Hooke’s lw (Eq. 1.3), constant of proportionately E is termed as the elastic modulus, and is the property of the material of the bar.
The above discussion can be seen in a slightly slightly different light. The deformation produced in the bar depends upon the load, the geometry of the bar and its material. So does elongation. elongation. But the geometry geometry of the bar is absent absent from Eq. 1.3. Once you convert the load to stress stress and deformation deformation to strain, the geometry of the structure becomes irrelevant.
2 Deformations, strains material properties
The above is a very powerful insight and forms the basis of the fundamental strategy of analysis in mechanics mechanics of deformable materials. materials. This strategy can be stated in the following manner: To determine deformations in a structure under a given loading, we first convert the loading to stresses using equilibrium considerations, convert stresses to strains using the material properties, and the use the geometry of the structure to determine deformations from the strains so calculated. This is also stated as a formula: macro to micro, conversion at micro level, and then micro to macro . Loading is macro, macro, stress and strain strain are micro level, and deformation is macro level.
and
We may, at time, need to go in the reverse direction. direction. We may be given the total deformation from which we need to determine the loading: we calculate strains from deformation, convert strains to stress using material properties, and the integrate stress to find the loading. The strategy here too is macro macro to micro to micro to macro.
2.1 Fundamental strategy of mechanics of deformable mechanics All structures resist loads by deforming. deforming. A structure deforms as a load is applied applied to it. As it deforms, the stresses build build up within the structure to resist resist the applied load. More the load more more is the deformation deformation and more are the stresses. The deformation increases till the resulting stresses are sufficient to balance the applied load.
Example 2.1 Tug of war Consider a tug of war in which 6 young men are pulling on a manila rope (crosssectional area: 6 cm2) with forces as shown. Find the net elongation of the rope.
We had considered in Sec. 1.2 the δ L / deformation of a uniform bar under a = longitudinal load. If we take take bars of various cross-sectional areas and of various lengths and plot the variations of deformation with the loads, we obtain plots as shown in Fig. 2.1b. ζ =P/A P P The variations are largely linear with slopes that are different for different (a) (b) (c) bars. We had seen earlier that if we Fig. 2.1 Elongation of bars with loads convert the load to stress (stress, ζ = load, P /cross-sectional area, A) and deformtion to strin (strin, ε = elongtion, δ /original length, L), the lines with δ
Area A Area A
ε
l
δ
17
By simple balance of forces we obtain T AB as 250 N. Similarly, from the other FBDs we get T BC = 500 N, T CD CD = 800 N, T DE = 550 N, and T EF = 300 N. If we assume a uniform distribution of stresses, which will be quite true away from the points where loads are applied (St. Venant principle), we can find the stresses in each section by dividing its tension by the cross-sectional area of the rope. Thus, the stress stress in section section AB is ζ = 250 N/0.0006 m2 = 416.7 kPa. The stresses in other sections can be determined in a similar manner and are given in column 3 of Table 2.1 below. Stresses to strain:
Stresses nd strins re relted by Hooke’s lw: ζ =Eε, where E is the elastic modulus of the material. A search of literature reveals significant variations in the value of E for manila ropes. A value of 100 100 MPa appears to be a good approximation10. We get the strains by dividing the stress values values by this value of E as shown in column 4 of Table 2.1.
Fig 2.2 Tug of war
Solution: Each section of the rope has a different tension, resulting in different stresses, strains and elongations. Let us take the four steps of our strategy in sequence.
Table 2.1 Calculation of elongation of the rope of Example 2.1
(1) (2) (3) (4) Section Tension, N Stress, kPa Strain AB 250 416.7 4.16×10- 3 -3 BC 500 833.3 8.35×10 CD 800 1333.3 13.33×10- 3 -3 DE 550 916.7 9.20×10 -3 EF 300 500.0 5.00×10
Loading to stresses:
To find the stresses, we first need to determine the tension in each section of the rope. There are five distinct sections sections carrying tensions. (The sections sections at either end do not have any tension, and, therefore, do not need any consideration.) This is done by making appropriate FBDs which will make the required tension force
(5) (6) Length, m Elongation, m 1.5 6.24×10- 3 -3 2.0 16.70×10 1.5 20.02×10- 3 -3 1.5 13.78×10 -3 2.0 10.00×10
Strain to deformation:
Once the strains are calculated, we can determine the elongation in each section by multiplying the strains with the length of the respective respective section. This has been -3 shown in column 6 of the table. The total elongation is 66.74×10 m, or 6.67 cm.
Example 2.2 A hanging cable Consider a cable cable of uniform section section hanging as shown in Fig. 2.4. A steel cable hangs from a roof under under its own weight. It is important to see that that the tension along the length of the cable is not constant, but varies. This can be seen by considering two sections, one at level b, and the other at level c. We have, in Fig. 2.4b, drawn the FBD of the cable up to level b. The tension T 1 is clearly equal to weight W 1. The weight W 2 in the FBD of Fig. 2.4c is no doubt greater than W 1, and therefore, T 2 is greater than T 1. Thus, the tension, and hence the stress along
Fig. 2.3 FBDs for determining tensions in the rope
an external force. Fig. 2.3 shows a sequence sequence of FBDs drawn for this purpose. The first of these has externalised the tension T AB in the section AB of the rope.
10
18
E for the rope accounts also for its unraveling as it stretches. The value of E
the cable increases from bottom to top. The diameter of the cable is dictated by the maximum stress in the cable. There is a considerable wastage of material, since the material near the bottom is not loaded to its capacity.
order in dx.
, neglecting terms of second
The force balance will then give, on simplification:
, where the negative sign indicates that diameter D decreases
as x increases. Using the boundary condition condition that the diameter D = D0 at x = 0, 11 we can solve this to get
In situations where the material costs are heavy, it often pays to reduce the cable diameter as we go down from the top. Fig. 2.5 shows a cable where it has been done. Its diameter at x = 0 is Do, and at x = L is D1. The variation in diameter is (a) (b) (c) such that the stress at any given location is constant equal to ζ o. Find the Fig. 2.4 A hanging cable differential equation governing the variation of diameter with x. Also set up an equation to determine the total total extension of the cable as a function of the various parameters involved.
How would we now calculate the total elongation of the cable? We first convert the stress at each section to the strain at at that section. Since the stress everywhere is the same, equal to , the strain is also the same everywhere equal to , where E is the elastic modulus. modulus. And, therefore, therefore, the total elongation is easily determined as .
Example 2.3 Deflection in an elementary truss Consider again the truss with a roller support (Fig. 2.6) discussed previously as Example 1.3, where the forces and stresses in the various members AC , BC and AB were determined. It was shown that there is no force in AB, a tension of 28.3 kN in member AC , and a compressive force of 20 kN in the member BC . The respective stresses were found to be 0, 160.1 MPa, and 113.2 MPa. Determine the displacement of point C where the load of 20 kN is applied. applied. All members are made of steel ( E = 210 GPa).
Solution: What is important to realize in the problem is the fact that the tension at any location varies with x, the distance from the roof. The weight weight supported supported by the section at x reduces as x decreases. We first determine x. the tension as a function of x
Solution: The equilibrium part has has already been solved. So also the stress determination. determination. We now need to find the displacement of point C to C 1.
Fig. 2.5 Cable of variable diameter
To determine the tension at x, we take a slice of thickness dx of the of the cable at x, and draw its FBD. This FBD has three forces, the upward tension T at x, the downward tension T + dT at x + dx, and the weight d W of this slice of the cable. Clearly equilibrium requires that the algebraic sum of the three be equal to zero. If the variable diameter of the cable is denoted by D as a function of x, the tension T is easily estimated in terms of the stress ζ o as , the stress being given as constant throughout. The weight dW is easily estimated as , where ρ is the density of the cable material and g the acceleration due to gravity. gravity. To determine the tension T + dT , we have to note that the area at the bottom of the FBD is different from that at the top. The tension here is thus,
./
For this purpose, we first find the elongation (or shortening) of members AC and BC . The stresses in the two members have already been calculated. The strains E. Thus, are now calculated as stress/ ε AC = 160.1 MPa/ 210 GPa = 0.76×10-3, and ε BC = − 113.2 / 210 G = − 0.54×10-3,
[..// ]
11
What about the second boundary co ndition: D = D1 at x = L? There is no other constant to determine. Or is there not? The resolution of this question lies in the fact that the constant stress ζ 0 cannot be specified independently of D1 and L. How would you determine ζ 0 in terms of the other parameters?
19
the negative sign indicating that the strain is compressive. We convert the strains into elongations. The lengths of the two bars are 1.41 m and 1 m, respectively, and the member AC elongates by δ AC = (0.76×10 3 )×1.41 m = 1.07×10-3 m (or 1.07 mm), and the member BC by δ BC -3 = − (0.54×10 )×1.0 m = − 0.54×10-3 m (or 0.54 mm).
A
a perpendicular to AC at E (to represent the tangent to the arc through E . The point where the two perpendiculars meet is the location of C 1. 1. We express the new location by specifying the x- and y- displacement of the point C. The xdisplacement is easy to determine. determine. It is nothing but EC , which is δ BC = 0.54 mm.
A
We note that the y-displacement is EC 1 = EF + FC 1 = CG + FC 1. But, CG = o CD /sin 45 = 1.07 mm/ 0.707 = 1.52 mm. we also see that FC 1 = FG = 0.54 mm. Thus, the y-displacement of the point C is 1.25 mm.
1m B B
C
E C
The procedure outlined here is quite a standard procedure for determining the displacement in pinned structures like this truss.
D
20kN
C 1
1m
2.2 Statically indeterminate problems
(b)
(a)
In the examples solved above we first obtained loads and stresses in the members before moving to the next step of determining strains and stresses. There are, however, problems where we cannot determine the forces in members without bringing in the considerations of deformation.
Fig. 2.6 Determining displacement in a truss
To find the new location of point C , we use the following following strategy: We change the lengths to the rod to their new length and make them rotate in arcs about their pivot points A and B, respectively. The point where the two arcs intersect is the deflected location of point C . Fig. 2.6b shows the geometric construction for determining the displaced location C 1 of point C. Bar AC has been shown elongated to AD (with CD being equal to AD as radius. The δ AC =1.07 mm. We draw an arc with centre at A and AD displaced point C should be on this arc. Similarly, we displace point point C to E by an amount δ BC = − 0.54 mm. We drw n arc with centre at B and BE as radius. The displaced point C should be on this arc as well. The point of intersection of the two arcs then represents the new location C 1 of C . Determining the intersection point of two arcs is a fairly lengthy process. Fortunately, since the changes in lengths are very small fractions of the lengths of the rods, and therefore the angles of swing are quite small, we can adopt an approximate procedure wherein we replace the arcs by the tangents to the arc. We draw a perpendicular to AD at D (to represent the tangent to the arc through D. Similarly, we draw
Consider as an example a beam supported on two ends as shown in Fig. 2.8. Since Fig. 2.8 A propped beam the bem’s deflection is excessive, it is propped up in the middle by a column. Also shown is the FBD of the beam. There are now three unknown reactions and only two equations (vertical force balance and moment balance) balance) to determine determine them. The problem is, therefore, statically indeterminate. It is easy to see that the reaction from the central prop depends on the rigidity of the beam. If the beam is quite rigid, the beam may not even touch the central central prop and there will will then be no reaction there. But as the rigidity of the beam decreases, the load borne by the central central prop increases. Therefore, it is necessary to bring in the considerations of deformation even to solve the problem of equilibrium. The general strategy for solving the problems of the mechanics of deformable bodies consists of three major steps:
Fig. 2.7
20
force it applies on the board be R2. Fig 2.7c shows shows the resulting FBD of the board.
Consideration of static equilibrium and determination of loads, Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations), and Considerations of the conditions of geometric compatibility.
Equilibrium analysis
There are two unknown forces R1 and R2 in this problem. There is also the unknown distance x that is to be determined. The equations of equilibrium give:
In statically indeterminate problems we cannot take these steps in a linear sequence, because there are not enough equations of equilibrium to solve for all the unknown loads. We have to consider simultaneously all three steps, even if we were interested in only one, say, in determining the forces in the system, as in the problem above.
∑ ∑ :
(a) (b)
We illustrate this strategy first with a simple example involving springs, and then with a little more involved problems.
The moments have been taken about about the location of the first support. There is no other equation of equilibrium. equilibrium. We cannot solve for three unknowns unknowns from these two equations. equations. This is why this problem is statically indeterminate. We nonetheless move on.
Example 2.3 A spring board
Load deflection equations
Fig. 2.9a shows a rigid board mounted on two similar springs, each of spring constant12 k, and length h. A man with weight P walks out on the board till he reaches a point a distance x from the second spring when the tip of the board touches the foundation platform as shown shown in Fig. 2.9b. What is the value of x in terms of P, L, h and k ? Neglect the weight weight of the platform.
Using the spring constant k :
(c) (d)
This introduces two more equtions, but two more unknowns, δ 1 nd δ2 as well.
Solution: It is clear from the geometry of the problem that as the man walks out, the first spring elongates applying on the board a downward force, while the second spring compresses applying an upward force on the board. Let the first spring elongate by an amount δ1 and let its tension be R1. Let the second spring shorten by δ2 and the upward
Geometric compatibility
From the geometry of the similar triangles in Fig. 2.9b, we can conclude that the length of the first spring after deformation will be twice that of the second spring after deformation .
(e)
The problem is solvable now. now. There are five equations and five five unknowns, R1, R2, δ1, δ2, and x, the distance to be determined. R2 from Eq. (a), calculating δ1 and δ2 from Eqs. (c) and (d) Writing R1 in term of R R2 only, and then substituting for δ1 and δ2 in Eq. (e), we can solve in terms of R for R2 to obtain
And then from (a): Fig. 2.9 A spring board
R1 and R2 in Eq. (c) we get the desired result, Using these values of R
12
Spring constant k is defined as the force required for producing a unit deflection in a spring. Thus, if a force F produces an extension (or a compression) of δ of δ, the spring constant is given by k = P/δ. Its unit is N/m.
21
Example 2.4 A bolt and a sleeve
L1 = L2 =5×10-2 m,
A brass sleeve of length 50 mm (ID: 14 mm, OD: 18 mm) is held between a nut and the head of a steel bolt of dia 10 mm (Fig. 2.10). The nut is brought flush with the sleeve and then tightened one quarterturn. If the pitch of the bolt is 0.7 mm, mm, determine the tension in the bolt.
E 1 = 110 GPa (for brass), and
The resulting stresses will be 225 MPa in brass sleeve, and 282 MPa in the steel bolt. Please note that these stresses are about the maximum that the sleeve and the bolt can sustain. Any more tightening would would result in failure of either or both.
Solution:
Example 2.5 A kinky spring
This is a statically indeterminate problem. As we tighten the nut, a compressive force builds up in the sleeve which shortens in length. Simultaneously, a Fig. 2.10 A bolt, sleeve and nut tension is built up in the bolt which elongates. Even though we are interested only in calculating a force, we cannot do so without bringing in the considerations of stress, strain and elongation.
Fig. 2.11 shows an elementary design of a spring which has one spring constant for the initial deflection (till the load is compressing only the inner aluminium tube), and then it increases when the load plate touches the outer brass cylinder so that the load deflection curve looks like that shown in the figure. For the inner inner tube of 40 mm dia and 2 mm wall thickness and the outer tube of 50 mm dia and 2 mm wall thickness, find the two spring constants.
E 2 = 200 GPa for steel.
Using these values, we get P = 225 N.
Equilibrium analysis
This is simple. If a compressive force P builds up in the sleeve, the tension in the bolt is P. Force deformation analysis
Solution:
The force deformation analysis is done using the macro-micro-micro-macro strategy developed in the previous section. To find the contraction in the length of the sleeve in terms of the compressive force P, we first find the stress ζ = P/A, P/A, the convert it into the strain ε = ζ/E = P/AE . The total contraction contraction then is δ1 = εL = PL1 /A1 E 1. The elongation of steel bolt is, similarly, δ2 = PL2 /A2 E 2.
Till the load is borne only by the inner aluminium cylinder
Fig. 2.11 Kinky spring
For a load P the stress in the cylinder is obtained by dividing the load by its cross-sectional area, which is πDt = π (40×10 (40×10- 3 m)(2×10- 3 m) = 2.5×10- 4 m2. -4 Therefore, the stress is P/ 2.5×10 2.5×10 Pa (compressive).
Geometric compatibility
We convert this stress into strain by dividing it by the elastic modulus for - 4 aluminium, which is about 70 GPa. Therefore, the strain strain is (P/ 2.5×10 2.5×10 9 -8 Pa)/70×10 Pa = 5.71×10 P (compressive).
-4
The nut moves through a total distance of ¼ of 0.7 mm, or 1.75×10 mm. If the steel bolt did not elongate at all, this would be the magnitude of δ1. But the bolt elongates through δ2. This means that that the contraction of the sleeve would decrease by an amount equal to δ2: Thus,
For a length of 300 mm, this strain results in a shortening of the cylinder by ζ×L -8 -3 -8 = (5.71×10 P)×(300×10 m) = 1.71×10 P m.
-4
δ1 = 1.75×10 m − δ2.
The spring constant13 is load per unit deflection, i.e., P/ (1.71×10 (1.71×10- 8P) = 58.3 MN/m
Substituting the values of δ of δ1 and δ2 in this condition, we get: PL1 /A1 E 1 = 1.75×10-4 m – PL2 /A2 E 2
Here,
A1 = π [(18×10 [(18×10-4 m)2 − 14×10-4 m)2]/4 = 1.00×10-6 m2, A2 = π (10×10 (10×10-4 m)2 /4 = 0.78×10-6 m2,
13
One can find the spring constant quite directly by noting that the stress is P/A, E, or P/AE. Deformation, then, is δ = PL/AE. The spring strain is stress/
22
The spring constant changes when the compression exceeds 0.1 mm, or when the 6 -3 load exceeds (58.3×10 N/m)×(0.1×10 m) = 5.83 kN
brass as 110 GPa, which is about in the middle of the range of values given in handbooks for brass or bronze as between 100 and 125 GPa.
After the outer cylinder kicks in
Putting in the values of the parameters in Eq. (c), we obtain: -8 -9 1.71×10 F 1 = 8.80×10 F 2 +
After the outer brass cylinder also starts bearing the load, the nature of the problem changing drastically. We no longer longer can use equilibrium alone to determine the loads loads borne by the two two cylinders independently. It is a statically indeterminate problem where considerations of equilibrium are interwoven with the consideration of deflections.
F = 0.51F + 1
or, (d)
2
14
Eqs. (a) and (d) can be solved simultaneously to obtain
Equilibrium of forces
and
.
Let us assume that after the total deflection exceeds 0.1 mm, the load borne by the inner cylinder is F 1 and that by the outer cylinder is F 2. Clearly, the equilibrium requires that: P = F 1 + F 2
(e)
Finding the spring constant for this part is a little tricky. Note that we should subtract 0.1 mm from the deflection δ and 5.83 kN from the load P to obtain the slope of the second part of the load deflection deflection curve of Fig. 2.11. We note that . The spring constant after the kink is given by:
(a)
There is no other equation of equilibrium that can help us determine the two forces. Force deformation considerations
The value of δ of δ2 in terms of F 2 is given obtained from Eq. (b) as δ2 = 8.80×109 F 2. Eq. (e) gives F 2 in terms of P. Thus, we we obtain
Let us move ahead and determine the stresses in the two cylinders, which are F 1 / A1 and F 2 / A2, respectively, where A1 and A2 are the two areas.
, or
A1 E 1 and F 2 / A2 E 2, both compressive, where The corresponding strains will be F 1 / elastic moduli. The contractions will be: E 1 and E 2 are the two elastic A1 E 1 and δ2 =F 2 L2 / A2 E 2. δ1 = F 1 L1 /
(b)
,
This is more than three times the value when only the aluminium cylinder was being deformed.
We have taken all steps according to the procedure outlined in Sec. 2.1 and illustrated in the examples there: load (macro) to stress stress (micro) to strain (micro) to deformation (macro). Geometric compatibility
2.3 Lateral strain: Poisson ratio
We know that the contraction of outer cylinder begins only when the inner cylinder has compressed compressed by a full 0.1 mm. Thus, the geometric compatibility requires that:
Consider a bar of length L subjected to an axial load P as shown in Fig. 2.12. Under the action of this load, the bar elongates by an amount δL so that an axial strain is set up equal to . From our experience experience with elastic materials materials like rubber bands we expect that as the bar elongates its cross-sectional area decreases, i.e., transverse strains ε yy and ε zz are also set up. Simeon Poisson, Poisson, a French mathematician in the early 19th century proposed that the strain in a
(c)
-4 2 The values of the various parameters are: L1 = L2 = 0.3 m, A1 = 2.5×10 m , A2 = 3.1×10- 4 m2, E 1 = 70 GPa, and E 2 = 110 GPa. We have used a value of E for
14
One can verify this result by using P = 5.83 kN, the value where the brass cylinder just kick in and finding that and as it should be.
L, A and E we constant, therefore, is k = P/δ = AE/L. Plugging in the values of L, obtain the same value as above.
23
transverse direction is a fixed (negative) proportion of the strain in the axial direction. Thus,
These are the tensile (or compressive, if the algebraic sign is negative) strains on the material due to tensile stresses. We can show using symmetry arguments that shear stresses do not cause any tensile strain. Therefore, Eqs. 2.2-2.4 can be used to convert stresses stresses into strain. These relations can be taken as extensions of Hooke law for uniaxial stress, and collectively are known as generalized Hooke law.
P
(2.1)
L
where ν is termed as the Poisson ratio and is a property of the material15.
Example 2.6 Compression of a block of rubber A block of rubber 50 mm×50 mm×30 mm is placed in a cavity and compressed with a force of 1 kN (Fig. 2.13). Determine the decrease in the thickness of the block.
z Next, consider the case where a y P second normal stress ζ yy is also x present. Because the stress – strain – strain relation is linear, the additional Fig. 2.12 Transverse strain under axial strains produced due to ζ yy will load simply be superposed on the strains due to ζ xx. Further, since the properties of the material are independent of the direction (termed as the property of isotropy), the strains due to ζ yy are:
Solution: As the rubber block is confined within a cavity, its transverse dimensions will not change and therefore there will be no strain in the y- and z-directions. The (negative) stress in the x-direction produces a (positive) strain in the transverse directions. The walls of the cavity prevent the rubber from expanding in that direction, and as a consequence, compressive stresses in y- and thus, a multimulti z- directions result. This is, thus, axial loading situation and the generalized Hooke law (Eq. 2.2-2.4) applies. Clearly, ζ yy Fig. 2.13 Compressing a rubber and ζ zz re equl (= ζ, sy) becuse of block symmetry. (we cannot cannot distinguish between y- and z- directions. The stress ζ xx is equl to (−1 kN)/ (50 mm× 50 mm) = −400 kPa, the negative sign indicating that it is a compressive stress. The value of E for rubber is about 500 kPa with the value of Poisson ratio ν of 0.5 of 0.5 (note that this value indicates that the rubber is essentially incompressible!). The three equations then give:
Similarly, the strains due to ζ zz are:
.
Since these three strains in the three directions due to three different stresses are all additive, the total strains in the three directions will be:
,
(2.2)
, and
(2.3)
.
(2.4)
-
15
It is interesting to explore what change in volume results when a uniaxial load δz in the is applied. Consider a small cuboidal element of dimensions dimensions δx, δy, and δz in x-, y-, and z-directions. If it is subjected to a load that causes an x-direction strain equal to ε xx, the resulting y- and z-strins re − νε xx each. The new dimensions of the elements now are, δx(1 δx(1 + ε xx), δy(1 δy(1 − νε νε xx), and δz (1 (1 − νε νε xx). Therefore, the deformed volume is δx•δy•δz • (1 + ε xx)•(1 − νε νε xx)•(1 − νε νε xx) ≈ δx•δy•δz •(1 + ε xx − νε νε xx − νε νε xx) = δx•δy•δz •[1 + (1 − 2ν) ε xx], neglecting terms of higher order in ε. The fractional change in volume is, thus, (1 − 2ν) ε xx]. If the material is incompressible, the vlue of ν, the oisson rtion must be ½.
,
(a) (b) (c)
Eqs. (b) and (c) are identical (as expected, because of symmetry), and from these ζ = −400 k. Using this vlue in Eq. (), we get we obtain ζ = ε xx = 0, i.e., no strain at all.
24
What is going on?
is obtained by dividing it by the cross-sectional area of the cylinder which sustains this load, i.e., 2 . The two stresses are superposed, superposed, and the net stress in the axial direction is given by
This strange result is the consequence of the value of Poisson ratio being taken as 0.5, which is the same thing as assuming the material to be incompressible (see footnote 5). Clearly, no strain can can be produced in an incompressible incompressible material confined in a cavity!
The negative sign is because the force F causes compressive stresses. The hoop stress is given by Eq. 1.8 as:
Let us redo this problem with with a different material. Let us assume that the the value E is 500 kPa and that of ν of E of ν is 0.4, the other conditions remaining remaining the same. The three equations now are:
, -
(a)
(d)
(b)
The force F makes no contribution to the hoop stress. The strain in the axial direction is given by Eq. 2. 4 as
(e)
(f)
As before, Eqs. (e) and (f) are identical, and from these we obtain ζ = −266.7 kPa. Using this value in Eq. (d), we get ε xx = -0.37, a very large strain. The thickness of the block will reduce by 37%.
Where ν is the Poisson ratio. Using the above values and the fact that ζ rr rr is of order p and, therefore, negligible, we get
Example 2.7 A confined pressure vessel
The value of ν of ν is always less than 0.5.
A long thin-walled cylindrical tank is snugly fitted between two rigid walls as shown in Fig. 2.14. It is is then pressurised to an excess pressure of p. If radius of the cylinder is R and its wall thickness t , determine the compressive force exerted by the wall on the cylinder.
2.4 Shear strain A material subjected to shear stresses deforms in shape. Fig. 2.15 shows shows an element of a material that is under the action of a shear stress. stress. We have seen earlier that shear shear stresses occur in pairs: If on the x-faces of an element there is are stresses η xy giving rise to a counter-clockwise couple as Fig. 2.15 Shear strain shown, there must occur on the complementary yfaces stress components that result in a clockwise clockwise couple. Please note that all the stresses shown here have the same +ve signs16. If the direction of one of the stresses was changed, the direction of all the components will be changed.
Solution: The length of the pressurized unrestrained cylinder increases because Fig. 2.14 A cylindrical tank between two walls of the combined action of the axial and the hoop stresses. This results in the cylinder trying to push the walls outward, which in turn apply a compressive force F on the the cylinder. cylinder. We can determine this force by noting that the net strain in the cylinder in the zdirection should be zero.
Though the element is under equilibrium under the action of these four types of stresses and the consequent consequent couples, the element deforms in in shape. It can be shown by some very interesting symmetry arguments17 that these stresses cannot 16 In fact, we have deliberately defined our sign convention for stresses to ensure this very situation. 17 The reader is referred to Crandall, Dahl, Lardner, Mechanics of Solids, McGraw-Hill, 2nd SI edition, Section 5.4 for these fascinating arguments.
There are two sources of axial stress ζ zz zz in the cylinder: one is due to the internal excess pressure ( from Eq. 1.9), and the other is the axial force applied by the walls on the cylinder. The compressive stress due to this force F
⁄
25
result in any linear (tensile or compressive) strains: the linear dimensions of the element will not change. It is only the shape that that changes in this case. The change in shape is manifested in the change in the included angle between two lines which were parallel to the axes. axes. Thus, angle at A reduces to α. We define the shear strain in the element at this point to be the change in angle, i.e., ( π /2 – /2 – α), for an infinitesimal element in the limit that the dimensions of the element tend to zero. The symbol used for for the shear strain strain in the x-y plane is γ xy, the Greek letter gamma. Thus,
. /
Solution: From the given coordinates, the length of the line
√
AB is . The original length of this line aligned with x-axis was 0.20 mm. This gives a contraction of 0.006 mm, – 0.006 mm/0.2 mm = − 0.03 or − or a strain ε xx of – 0.006 3%.
Similarly,
Fig. 2.16 Calculating strains
and adding to it the angle the line AD . Thus,
The shear strain, then, is
This is the shear strain in the material.
We next calculate the shear stress. From Eq. 2.6,
-
, or −150 , where
we have used 26 GPa for the value of G for steel.
We next calculate the linear stresses from Eqs. 2.2 and 2.3:
Shear stress in a plane depends on the shear strain in that plane alone, and not on any other component of strain, shear or tensile. tensile. Similarly,
:
(2.7)
:
Eqs. 2.6 and 2.7, together with Eqs. 2.2 – 2.4 form the complete set of strainstress relations.
Solving these two simultaneously, we get ζ xx = −5.57 G, nd ζ yy = −2.70 G. The negative signs indicate that both these stresses are compressive.
Example 2.8 Calculating strains and stresses
Example 2.9 A vibration isolator using rubber in shear
Fig. 2.16 shows a small element of a steel structure after deformation. The undeformed element was aligned with the axes and its dimensions were 0.2 mm × 0.2 mm. The coordinates of its vertices A, B and D after deformation are given as (in mm): A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). Clculte the vrious strains and stresses.
It will be shown in Chapter 6 that E , G and ν are related:
is
makes with the y axis (
The shear modulus of most materials is about a third of their elastic modulus18.
18
AD
where G is termed as the shear modulus of the material. The shear modulus has the dimensions of force/area or ML- 1 T- 2 . It has the the same unit as as E has, namely Pascal, Pa. Table B.1 in Appendix B gives some representative values of shear moduli and Poisson ratios of some common materials. Steel has a Poisson ratio of about 0.27. Aluminium and brass have a little higher higher values of 0.32 and 0.35, respectively. Soft rubber has a value of about 0.5, signifying that its volume does not change appreciably as it stretches.
and
line
(
(2.6)
of
To calculate the shear strain γ xy we determine the angle DAB. That is obtained by subtracting from π/ 2 the angle the line AB makes with the x axis
As for the tensile stress and strain, shear stress too is linearly related to the shear strain. ,
length
. The original length of this line aligned with y-axis was also 0.20 mm. mm. This gives a contraction – 0.004 mm/0.2 mm = − 0.02 or − 2%. of 0.004 mm, or a strain ε yy of – 0.004
This is dimensionless as the the linear strains were. The angles are measured in radians
the
√
(2.5)
Figure 2.17 shows the the schematic of a vibration vibration isolator. The machine whose vibrations are to be isolated is mounted on the central steel block and applies a vertical load of 8,000 N as shown. shown. It is suspended from two vertical vertical walls through two rubber pads of height 12 cm and cross-sectional area of 10 cm ×10 cm. The two rubber pads act as shear shear springs. Determine the total deflection deflection of the central post under the load and the effective spring constant of the two pads taken together.
26
8,000 N
Solution: Fig. 2.18 shows a pad in deflected configuration. It shares the total load of 8,000 N with the other pad, and therefore, carries 4,000 N as a shear shear load. This load acts on an 2 area which is 12 cm×10 cm, or, 0.012 m . The sher stress is −4,000 N/0.012 m 2, or, . With G for rubber as 1.0 GPa, the = −333.3 k/1.0 shear train is G = −0.33. The totl liner deflection, then, is −0.33×10 cm = −3.3 cm.
There is no change in shear strain because of thermal expansion, i.e., a shear strain arises only due to the respective shear stress component. If the structure in not permitted free expansion due to geometrical constraints, stresses may be setup in the structure. structure. Such stresses are frequently termed termed as thermal stresses .
Since a load of 8,000 N produces a vertical deflection of 3.3 cm, the spring constant of the two pads taken together is 8,000 N/0.033 m = 242.4 kN/m.
In examples below we illustrate how the thermal and elastic strains can be combined. Wall
Wall
Example 2.10 Thermal stresses
Rubber blocks
Fig. 2.19 shows a 20 cm long 3 cm dia steel rod held between two rigid supports. The temperature of steel rod is raised by 100 0C. What is the force in the bar?
Fig. 2.17 Schematic of a vibration isolator
Solution: The increase in temperature will tend to increase the length of the rod. But the rigid Fig. 2.19 supports will prevent it from elongating. This results in an axial force. If F is the compressive axial force acting on the rod A = F/ because of the walls , the stress ζ xx = F / 7.06×10−4 m2. Since there is nothing that is applying a force in the other two directions, ζ yy and ζ zz are both absent. Also, since there is no net change in the length of the bar, the axial strain ε xx is 0. Using these values in Eq. 2.9, we get:
2.5 Thermal Strains As the temperature of a body changes so does its dimensions. The change of length with temperature is generally quite linear Fig. 2.18 with changes in temperature of several hundred degrees Celsius. If we heat a rod of length L through a temperature change of ∆ , ∆T , its length changes as where α is termed as the coefficient of thermal expansion . This can be seen as producing a strain εT given by
( ) , or
0 Here α for steel is about 11×10 / C, E = stress is E = 210 GPa, and ∆T = T = 100 C. The stress
(2.8)
-
,
or
231 −4
MPa,
m2), we get
the axial force F as 163 kN, a fairly large force.
Example 2.11 Residual stresses in electroplating
These thermal strains are superposed on the elastic strains produced by loading of members. The total strain produced in the the presence of elastic strain is obtained obtained simply by adding the two strains. Thus, using Eqs. 2.2-2.4, and Eq. 2.8, we get:
The corrosion resistance of steel is improved by electroplating it with a thin layer of copper, usually only a few hundred microns microns thick. For bright electroplating the bath temperature should be around 50 0C. After electroplating when the plated steel Fig. 2.20 Copper plating on steel steel returns to the room temperature, say 0 20 C, the higher coefficient of thermal expansion of copper compared to that of steel results in copper tending to
(2.9) , and
0
compressive. By multiplying this stress with the area of the bar (7.06×10
The unit of α of α is per . Representative values of α of α for some materials are given in Table B.1 in Appendix B. It should be noted that temperature changes do not produce any shear strains.
( ) ( )
.
-6
(2.10) (2.11)
27
contract more than that what the steel does. The bonding between steel and copper prevents this, and as a result steel does not let the copper layer contract more than it does itself. Steel layer ends up applying a tension to the copper layer, and the copper layer pulls the steel inwards, the total tension force in copper being equal to the compressive compressive force in steel. However, since the steel layer is an order of magnitude thicker than the copper layer, the stress in steel will be negligible compared to that in copper.
sleeve should be heated for fitting, and (b) the resulting interfacial pressure on cooling.
Solution: It is convenient to work in the cylindrical polar coordinates, r , , and z, as shown. The relevant equations for strains and stresses are obtained by simply substituting r , , and z for x, y, and z in the equations developed so far20.
19
Thus, there is a residual tensile stress in the copper layer. layer. Estimate this residual residual tensile stress.
To determine the temperature beyond which the sleeve should be heated for fitting, we use Eq. 2.8 with εT = 0.1 mm/20 mm = 0.005, and αsteel = 11×10- 6 / 0C, to get
Solution:
There is no stress at 50 0C. When the temperature temperature decreases to 25 0C ( ∆T = −25 −25 0 C) the net strain in the copper layer will be exactly the same as the net strain in the steel plate. The steel plate will have only the thermal strain, since we can assume that the thickness of the steel plate is so large (compared to that of copper film) that the same force in the t wo layers (equilibrium requirement) will result in negligible stress in steel. Thus,
. / ( ) Using the property values: , we get:
=
0
11×10- 6 / C;
-
455 0C
The sleeve has to be heated through a range of more than 455 0C to slip it on the shaft without forcing it. Now on cooling, the ID of the the sleeve must be equal to the dia of the the shaft. Let us look first at the stress and strain strain in the shaft. If the interfacial pressure is p, the radial stresses in the shaft are of order p. There is no other stress in the shaft. On the other hand, the sleeve will be subjected to a radial stress of the order of p and a hoop stress (radial stress ζ (refer to Section 1.8). The hoop stress is given by pR/t . Since R >> t, the hoop stress, and consequently, hoop Eq. 1.8 as strain are much larger larger than the compressive compressive strain of the shaft. We can, thus, neglect the compression of the shaft diameter in co mparison to the increase in the ID of the sleeve. Therefore, we can take the final dia of the shaft to be 20 mm itself, and the final ID of the sleeve too as 20 mm. As the dia of the sleeve increases from 19.9 mm to 20 mm, its circumference increases from π ×19.9 ×19.9 mm to π ×20.0 ×20.0 mm, giving a hoop strain strain of 0.005. Since the axial stress is absent, E . and the radial stress is of order p, much less than , we get 0.005 = / Using the value of E as 210 GPa, and expressing as pR/t , we get p, the interfacial pressure as 157.5 MPa
0
16×10- 6 / C; and
(11 – (11 – 16) 16) ×10- 6 / 0C×(− 250C)
= 15 MPa (tensile), not an insignificant stress
Example 2.12 Shrink fit A steel shaft (20 mm dia) is to be fitted with a steel sleeve (20 mm long, OD 26 mm). In order to tightly fit the sleeve, its internal diameter is made as 19.9 mm. mm. It is fitted on to the shaft by heating it till its internal diameter is larger than 20 mm, slipping on, and then Fig. 2.21 A shaft with shrinkletting it cool. As the sleeve shrinks, it presses presses fitted sleeve on to the shaft creating an interfacial pressure which results in a good bond. Determine (a) the temperature temperature beyond which the
2.6 Tensile test Tensile test is one of the primary tests that are used to determine experimentally the properties of matter. A standardized specimen (Fig. 2.22) which which uses a strip or a rod of uniform cross-section between two enlarged ends is gripped and a
20
This is possible since none of the equations we have developed so far has does partial derivatives of physical quantities with respect to x, y, and z. Since does not have the dimension of length, we cannot simply replace derivatives with respect to y by those with respect to , and the equations that have partial derivatives change in a more complex manner.
19
This residual tension can be very detrimental when the structure is subjected to tensile (or bending) bending) load. The copper layer layer would crack and and the corrosion resistance will be lost.
28
termed as the elastic limit. It is not possible to determine either the proportional limit or the elastic limit with any degree of accuracy.
tension force force is applied. An extensometer extensometer (Fig. 2.23) is used to measure how the length between two specified points on the uniform portion changes as the tension is increased. The tension is converted into stress ζ by dividing it with Ao, the cross-sectional area21, and the elongation is converted to linear Fig. 2.22 Tensile test specimens strain ε by dividing the elongation by Lo, the gauge length (the undeformed distance between the two points points on the specimen). On plotting the two we we get a graph like that shown in Fig. 2.24 for ductile materials like steel. The initial part of the curve is a straight line and represents the linear elastic range22 of the material. If a material is loaded to point A within this range as shown, and then unloaded, the material will return to its unloaded state (strain will return to zero). We say that if the material is loaded and unloaded within the elastic range, the material exhibits Fig. 2.23 Schematic of an extensometer used in a no permanent set . The slope tensile test of the stress-strain curve in the linear elastic range is E, the elastic modulus of the material.
The strains in the elastic region are very small, quite insensible without instruments to amplify their magnitude. That is why extensometers are used which amplify the strain, either mechanically or optically. When the stress in steel is of the order of 400 MPa, the relative displacement Fig. 2.24 Typical stress-strain curve for of two points originally 100 mm apart a ductile material would be of order of 0.2 mm. Strains in most structural materials are of the order of a few tenths of a percent at most. As the load is increased further the material stops showing the elastic behaviour, that is to say that if the material is loaded beyond a certain point and then unloaded, it will not return return to its original unloaded state. We say that the material now has a permanent set . In Fig. 2.25, the material material is loaded up to point Y. When the load is now decreased, the material follows the straight line path downwards as shown. The straight line has the same same slope E as the original linear elastic line, but now it meets the strain axis at a point left of origin, signifying that there is now a strain even when the loading and stress have been reduced to zero. The material is said to have yielded and undergone plastic deformation.
The stress level up to which the linear elastic range extends is termed as the proportional limit.
It is quite difficult to decide experimentally the point at which the plastic deformation begins. It is for this reason that we, quite arbitrarily, designate the point Y, where the permanent set (the residual strain when unloaded after loading to that point) is 0.2%, (or a strain of 0.002) as the yield point of the material. The corresponding stress ζ Y Y is termed as the yield strength of the material.
As the stress is increased, non-linearity comes into play, but the material still shows elastic behaviour, i.e., on unloading the material exhibits no permanent set. The stress level up to which the elastic range (linear or non-linear) extends is 21
Note that the stress here is only a nominal stress since we are not accounting for the changed changed cross-sectional area as as the specimen elongates. This is not significantly different from the true stress stress till the strain becomes quite large. In fact, the decrease in the stress at very large strains as shown in Fig 2.23 arises only because we are not using the true true stresses. The load does not decrease as strain increases. 22 The linearity and elasticity are two different different phenomena. The elasticity refers to the fact that there is no permanent set, i.e., the material on unloading returns to the original unloaded state, along the same curve it followed while loading. loading. This may or may not be linear. In fact, a rubber band is largely non-linear elastic.
Consider a material that is loaded Fig. 2.25 Yielding and permanent set beyond the yield point Y to a stress level Y 1 (> Y ). ). If the material is unloaded now, it will follow a straight line down from Y 1 with a slope of E to point B, and there will be a permanent set as shown shown in Fig. 2.26. If, after this
29
initial yielding and unloading, the material is loaded yet again, it will follow the straight line up from point B (with the slope E ). ). This material is now not going to yield again till it reaches the stress level of Y 1, higher than Y. Thus by loading the material to a level higher than the yield strength Y , we have been able to raise the effective yield strength of the material (in subsequent loadings) Fig. 2.26 Strain hardening to Y 1. This phenomenon is known as strain hardening and is frequently employed to increase the wear wear resistance on materials. Shot peening and sand blasting are examples of processes employed to strain harden structures. Cold rolling of structural steel also produces usable work hardening.
Beyond ultimate stress, a neck develops in the specimen; the area decreases locally while the nominal stress decreases. Ultimately, the fracture takes place at the neck. We have discussed so so far the stress-strain curve for for a ductile material. The corresponding curve for a brittle material is significantly different from that for a ductile material. Glass, cast iron, concrete, carbon fibre, and Perspex are some of the brittle materials. Many material ductile at room temperature may be brittle at low temperatures.
Most of the differences between the behaviour of ductile and brittle material arises from the fact that there is little plastic deformation in a brittle material. Fig. 2.28 shows the stress-strain curves for glass and cast iron. The first thing to note is that both these materials undergo fracture at very little strain, much less than that for a ductile material. Glass, for example, shows no plastic deformation at all.
Loading of the bar beyond the yield strength produces large deflections with small increases in the stress. The maximum stress that the material is able to attain is termed as the ultimate stress . When a material is strained beyond the ultimate stress, the stress appears to be decreasing as strains strains increases. As noted earlier this is a false impression, since what we are plotting here is the nominal stress obtained by dividing the load by the initial area Ao. The area of a ductile specimen decreases Fig. 2.27 Necking of ductile significantly when it undergoes plastic materials deformation. This is, usually, a local phenomenon. The area of cross-section of the specimen specimen suddenly starts decreasing at some point where there are imbedded dislocations in the material. This is termed as necking (Fig. 2.27). As the necking progresses, the true stress at the section increases and the specimen fractures.
Another feature Fig. 2.28 Stress-strain curves for some brittle materials to note is that the strength to fracture is much larger in compression than in tension. This is a usual characteristic of brittle materials.
2.7 Idealized stress-strain curves The stress-strain curves for the various materials do not admit any simple mathematical description. description. However, we do need such descriptions for analysis of structures. It is for this convenience convenience that we model the the materials into a few classes and use simple stress-strain relations to model the actual behaviour.
We summarize here the sequence of major events in the tensile test of a ductile specimen:
Linear elastic deformation till the proportional limit: . Non-linear elastic deformation till elastic limit. Yield point, where the permanent set is 0.2%. If we unload the material beyond yield point, and then reload it, the yield strength increases. This is termed as strain hardening. As the material undergoes plastic deformation, the stress level increase up to ultimate stress.
Most problems of high school mechanics are solved by assuming the structural members to be rigid. This means that there is zero strain for any level of loading. Fig. 2.29a depicts the stress-strain curve for a rigid material.
30
The stress-strain curve for a perfectly elastic material is as shown in Fig. 2.29b, a sloping straight line whose slope is the value of the elastic modulus E . We have used this idealization of materials in this book so far. This is a very useful model
The strain hardening model is shown in Fig. 2.29e. Note that the elastic-perfectly plastic graph of Fig. 2.29c cannot model the strain hardening phenomena, since the slope of the the plastic part is zero. Reloading the material material after unloading following some plastic deformation will not increase the yield strength. But the non-zero slope of the plastic part of the curve of Fig 2.29e ensures that the yield strength increases on re-loading after the material has been de-stressed after it has undergone some plastic deformation.
Example 2.13 Fitting an idealized curve The solid line in Fig. 2.30 shows the stress-strain curve of cold-rolled 1020 mild steel. Find a simple mathematical relation relation for load vs. elongation of a 20 cm long 20 mm × 20 mm cross-section bar.
Solution: A look at the actual stress-strain curve suggests that though the material exhibits a bit of strain hardening, adopting an elastic-perfectly plastic idealization will not be bad at all. We first determine the yield strength of this material by drawing a line parallel to the elastic portion of the curve from a permanent set value of 0.2% or from strain ε = 0.002. The light broken line so drawn meets the curve at a stress of about 580 MPa, which may be taken as the value of yield stress ζ Y Y for this material. Therefore, the idealized curve looks like like the broken heavy line shown in the figure. In other words, we may model 1020 CR mild steel as elastic up to a stress level of 580 MPa with a value of E E = 580 MPa/ 0.003 = 193.3 GPa (0.003 being the strain at the yield point). point). Thereafter, the steel may be taken to to deform perfectly plastically.
Fig. 2.29 Idealized stress strain curves
and is used extensively in most structural analysis, because most structures are designed to work within the elastic range. Another commonly used idealized model is the elastic-perfectly plastic one shown in Fig. 2.29c. in this, we assume that once the material yields, the stress remains constants at the yield strength strength level. We shall have occasion to use this in a few problems.
A stress of 580 MPa corresponds to a load of 580 MPa×(20×10-3 m)2, or 232 kN. In this region, the elongation will be δ = PL/AE = -3 2 P×0.2 m/(20×10 m) × (193.3 GPa) = 2.59×10 9 P m, where P is the load in Newton. Newton. We draw a line through the origin up to the yield point with P = 232 kN. After this, the deflection increases Fig. 2.31 Load-deflection curve without any increases in the load. Fig. 2.31 graphs this load load vs. elongation behaviour.
Fig. 2.29d shows the behaviour of a perfectly plastic material. In such a material there is no strain till the stress reaches a critical level, after which there is no increase in stress as the material deforms. This idealization may be used when Fig. 2.30 Stress-strain curve for cold rolled 1020 mild steel there are very large plastic deflections, so that the elastic-strain region of the stress-strain curve can be neglected.
31
tendons, while there is equivalent equivalent compression in the concrete. concrete. This residual compression in concrete helps increase the tensile load capacity of the concrete.
2.8 Pre-stressing Many brittle materials have much higher yield strengths in compression than in tension. For example, concrete concrete has a tensile strength of only 10-15 % of its compressive strength. It is common to use reinforcement in concrete to give it increased strength in tension. This is used quite effectively in the design of beams. We had seen in Section 1.3 that in a beam 2.32 Reinforcement in a concrete beam subjected to bending under vertical loads, its upper layers are in compression while the lower layers are in tension. It is common to increase the tension strength of concrete by reinforcing reinforcing it with tendons of high tensile-strength material, such as steel rods (Fig. 2.32).
Example 2.14 Pre-stressed concrete Let us consider a simple pre-stressed concrete structure to serve as an illustration of the calculations involved. involved. Let a concrete beam of of cross-sectional area 5 cm×5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after after the concrete is set. What is the residual compressive stress in the concrete?
Solution: 2
A tension of 200 kN in the steel bar results in a stress of 20 kN/[¼π kN/[¼ π (0.010 (0.010 m) ] = 255 MPa. (This is quite close to the yield strength of commercial structural steel.) -3 The strain under this stress is ε = 255 MPa/ 210 GPa = 1.21×10 , and the total elongation of the bar will be (1.21×10- 3)×2 m = 2.42 mm.
But many a times this is not sufficient. sufficient. One of the very interesting ways to overcome this problem is pre-stressing. If we build up within the concrete some residual compressive stresses when it is being cast , , subsequent tension loading of the structure will first relieve the compressive stresses before generating tensile stresses. In this way, the structure will be able to support enhanced tensile loads before the tensile stresses reach the critical level.
After the tension is released, the steel rod shortens, but it does not reach its unstrained length, and there is a residual tension in the steel tendon and an equalmagnitude residual compression in the concrete. This force cannot be determined by statics alone. We have to make recourse to deformations and apply a geometric compatibility condition. If δs represents the final elongation of the steel bar (from its unstrained condition, and δc represents the contraction of the concrete, it is clear from the construction of Fig. 2.34 that δs + δc = 2.42 mm. This is the geometric compatibility condition.
(a) Tendon as cast (b) The no-stress length of the tendon (c) The final configuration on release of tension Fig. 2.34
Let F represent the force, tension in steel and compression in concrete. concrete. Then, δs Ac E c = F (2 = FL / (2 m)/[π/ m)/[π/ 4(0.010 4(0.010 m)2]×210 GPa, and δc = FLc / (2 m)/(0.05 s A E s s = F (2 m×0.05 m)×20 GPa.
Fig. 2.33 Pre-stressing
Fig. 2.33 explains how the pre-stressing works. works. Tendons, most often steel wires, are put under external external tension and the concrete cast around around them. After the concrete is set, the external tension is released. released. The tendons now try to shrink to their no-stress lengths, but the concrete around them prevents them from shrinking to their original length. length. In the process, there is a residual tension in the
Thus, from the geometric compatibility condition we get,
,
32
Which gives F = 15.03 kN. This is the residual force which which produces a compressive residual stress of 15.03 kN/(0.05 m×0.05 m×0.05 m) = 6.01 MPa. This leads to considerable strengthening of the structure in tension.
forces, F 1, F 2, F 3, …, F n. Let the total total strain energy be U , which is a function of F 1, F 2, F 3, …, F n: .
Let us first load the body with all the forces F 1, F 2, F 3, …, F n, and, then, increase one of the load, say F i, by a small amount, say δF i. The additional strain work done can be written as:
2.9 Strain energy in an axially loaded members When a force F is slowly applied to a deformable body (that is supported in such a way that no rigid-body motion is possible) work is done as the body deforms. The work done U in deformation is
∫
(2.12)
Let us next reverse reverse the order of loading. We first apply the small load δF i on the body before any Fig. 2.37 An elastic body other load is applied. After this load is applied, F all other loads F 1, F 2, F 3, are applied on the body. Under the combined combined …, 3, n action of all these loads, the point of application of load F i moves through an in23 line distance of δ of δi. The work done by this small force δF i is then
Fig. 2.35 The area under the force-displacement curve is the work done in deformation
This is represented by the area under the F-δ curve of Fig. 2.35. For purely purely elastic deformations, the force-deformation curve is linear and therefore the work done in deformation is given by
dU = δF iδi
. This work is stored stored within the body
as mechanical energy and is termed as the strain energy. It is a conservative energy and is released when the load is removed and the body returns to its original configuration.
(2.13)
(2.14) 24
This is known as Castigliano theorem . It states states that if the total strain energy of a structural system is expressed in terms of the external loads applied to it, the deflection at any one loading point (in-line with the load ) is obtained by taking the partial derivative of the strain energy with respect to the load at that point. We illustrate the use of this theorem, first to two simple cases in Examples 2.15 and 2.16, and then apply it to a more involved involved problem of a truss. Example 2.17 introduces a process that can be generalized and can also be used for solving statically indeterminate cases as illustrated in Example 2. 18.
Fig. 2.36 A bar under an axial load P
Let us calculate the strain energy of a bar subjected to an axial load as shown in Fig. 2.36. The deformation δ (elongation, in this case) is given by FL/AE , and the strain energy U is then
(b)
Since the order of loading (in the linear elastic case) should be immaterial, we can equate the two values of work given by Eqs. (a) and (b), to get:
Let a number of forces F 1, F 2, F 3, … ct on the body (or the structure) and the resulting displacements at their points of application be δ1, δ2, δ3, …. The work done in deformation is independent of the order in which the forces are applied. We apply the forces simultaneously such that the total work of deformation is given by the sum of the work done by individual forces:
∑
(a)
Example 2.15 An elementary truss
.
Consider again the truss with a roller support (Fig. 2.38) discussed previously as Example 2.3. Determine using the energy method method the displacement of point C
2.10 Calculating deflections by energy methods : Castigliano theorem
23
The factor of ½ is missing in this expression because the load δF i remains constant while the displacement δi takes place. 24 There are some simple simple extensions to Castigliano Castigliano theorem. One of the simple extensions is its application to non-linear systems where the strain energy is replaced with , which is termed as the complementary energy.
Consider an elastic body which is fully supported in the sense that its rigid body translation and rotation rotation are not permitted. Let it deform under under the action of n
∫
33
∫
where the load of 20 kN is applied. All members are made of steel E ( = 210 GPa) and have a circular cross-section of 20 mm dia.
same as was obtained in Example 2.3. To determine the horizontal displacement at C, we take the partial derivative of U with respect to Q:
Solution:
Since the energy method involves taking the derivative with respect to the force applied at the point where the deflection is to be calculated, we replace the load of 200 kN with a variable load P. Also, since the use of Castigliano theorem gives displacement only in line of the force with respect to which the derivative of the energy is taken, we cannot find the horizontal displacement of point C since there is no horizontal force present there. However, we can overcome this deficiency by introducing a dummy force Q in the horizontal direction at C , and then taking the derivative with respect to Q before equating Q to zero.
,
Example 2.16 Tug of war revisited Consider once again the tug of war discussed as Example 2.1. We determine here the total elongation of the rope using the energy method. The cross-section of the 2 rope was given as 0.0006 m , and the value of the elastic modulus for the manila rope was taken as 100 MPa.
Solution:
A
from the vertical force
The amount of work involved here with the energy method is far less than that in Example 2.3, and the results here are more accurate.
A simple equilibrium analysis of the joint at C using the FBD given in Fig. 2.39b, gives:
, which on substitutions of values gives
the negative sign indicating that it is in a direction opposed to that shown for Q.
balance, and
from
the
1m
horizontal force balance.
These give
B C
, and
20kN
. The force in the element AB
is zero as before. The total strain energy, then, is:
( ) )
1m
Fig. 2. 38 An elementary truss Fig 2.40 Tug of war
To find the total elongation by energy method, we need to anchor the rope at one end (say, at point A) and apply a force P at the other end (see Fig. 2.41). The total energy will be determined in terms of P and partially differentiated with
The vertical displacement of point C (in line with force P) is obtained by taking the partial derivative of U with respect to P:
Replacing P with 100 kN, Q with 0, LCA with m, LCB with 1 m, E with 2 210 GPa, and A with π (0.02 (0.02 m) /4, we get
= 0.0058 m, nearly the
Fig. 2.41 FBD for application of energy method (a) Truss
respect to P. To find the total strain energy of the system, we need to determine the load in each section of the rope which can be obtained by simple equilibrium
(b) FBD of pin C
Fig. 2.39 Introducing forces P and Q
34
analysis. Once we obtain the forces we can calculate strain energies by evaluating . Table below organizes these calculations:
Solution:
As before, we need to introduce dummy forces P and Q, respectively, at points F and B in-line with the displacement displacement desired. The method of solution solution is exactly the same as in the previous examples, but because of the larger numbers of members involved we devise a scheme that will permit reduction of labour.
Table 2.2 Calculation of elongation of the rope of Example 2.16
Segment EF DE CD BC AB
Length, L
Force, F
2m 1.5 m 1.5 m 2m 1.5 m
P P + 250 N P + 500 N P + 200 N P − 50 N
2
F L 2
2P 1.5P 1.5P2 + 750P 750P + 93750 1.5P 1.5P2 + 1500P 1500P + 375000 2 2P + 800P 800P + 80000 1.5P 1.5P2 − 150P 150P + 3750
∑
4P 3P 3P + 750 3P 3P + 1500 4P 4P + 800 3P 3P − 150
The total strain energy of the truss will be the sum of the strain energies of the individual members:
∑
The total strain energy is given as , where summation is taken over all the segments of the rope. The deflection at the free end is obtained by taking the partial derivative of the total energy with respect to P.
The deflection in the direction of force P is obtained by taking the partial derivative of U with respect to P:
∑ ∑ ∑
∑ ∑ ∑ ∑
where the value of
(a)
Attention is drawn to the fact that the here are the forces in the members in the presence of all loads, including P and Q. The values of may be imagined to be made up of two parts, one due to the actual loads on the structure, and the other due to the application of dummy loads P and Q. After we evaluate the sum specified in Eq. (a) above, we plug in the value of P and Q as zero. This implies that in the final step, the value of ’s will be the original values of ’s without P and Q.
has been substituted substituted from Table 2.2. Plugging in the
2 A = 0.0006 m and that of E E as 100 MPa, and the value of P as 300 N, value of A we get m, or 6.67 cm, the same as was obtained in Example 2.1.
Example 2.17 Unit force method Consider the truss shown in Fig. 2.42. The truss is pinned at point A and is supported on rollers at point B. This implies that while at A, there could be horizontal as well as vertical reactions, there is no horizontal reaction at B. All joints are assumed as pin joints, and as has been explained earlier, all members are two force Fig. 2.42 Example 2.17 members and hence carry only axial loads, tensile or compressive. We will illustrate the use of energy method to determine the vertical displacement at joint F and the horizontal movement at the roller support B. All members are made of steel and have a cross-sectional are of 6 cm2.
The term
, which
Fig. 2.43
is the rate of change of with P can be interpreted as the additional force in the ith member due to a unit load at P. This interpretation is valid because of the th linear dependence on P of the additional load in the i member. We, therefore, organize our calculations in two distinct parts: We first calculate the actual values of ’s, i.e., the vlues of o f ’s without P and Q. These are used as values for the first part of the terms in Eq. (a). We, then, calculate the values of ’s tht would result when unit force is pplied in plce of P, without any of
35
−8
m2)×(210 GPa) GPa) = 1.588×10 all the members of the truss.
the other external forces being present. These values of ’s will be used in plce of the partial derivatives
.
− N 1. Table 2.3 gives the values of F and L/AE for
Calculation of through calculation of the F i’s for a unit load of P
This procedure is termed as the unit force method.
Refer to Fig. 2.45 which shows the FBD of the truss for a unit load at the location of P and no other load. Clearly, the reactions at A and B are 0.5 N, each. We again go from pin to pin, draw Fig. 2.45 FBD for unit load at F the relevant FBD and calculate the force in each member. th These have been shown in the 4 column of Table 2.3. Fig. 2.44 Some FBD’s for determining the forces
Determination of
Table 2.3 Calculation of vertical movement of pin C
’s in absence of forces P and Q:
Member Force L/AE −8 −1 Fi, kN (×10 N )
We first determine the reactions R1 x, R1 y, and R2 y by considering the structure as a whole bin the FBD shown in Fig. 2.44a. T he equilibrium equations are:
−5
∑ ∑ ∑ :
(×10 m)
,
:
AD −16.7 AF + 13.3 BE −25.0 BF + 20.0 −16.7 CD CE −16.7 CF +5.0 DF 0 −8.3 EF
, and
:
.
The three equations can be solved simultaneously to give and .
,
,
Once the reactions are known, we can go from pin to pin to calculate the axial forces in all the the members. Consider, for example, example, the FBD of pin B shown as Fig. 2.44b. By writing the two force balance equations and noting that , we get
∑ ∑ :
:
Solving the second of these first, we get
1.24 1.59 1.24 1.59 1.24 1.24 1.59 1.24 1.24
−0.67 0.89 −0.67 0.89 −0.67 −0.67 1 0 0 Total
,
The last column of Table 2.3 shows the product
.
, and then
13.9 18.8 20.8 28.3 13.9 13.9 8.00 0.00 0.00 117.6
for each member of
the truss. The sum of these over all the members gives the deflection of point D, in-line with the dummy force P. Thus, Δv,E = 1.176×10−3 m, or 1.18 mm.
.
We can similarly move from pin to pin and calculate all the forces.
This is quite a general procedure for trusses.
L/AE for each member of the truss. For example, We next calculate the value of L/AE L/AE is given as (2 m)/(0.0006 for member AF which is 2 m long, the value of L/AE
Horizontal deflection at the roller support at B:
36
We next calculate the horizontal deflection at the roller support at B. For this purpose, we place a dummy horizontal load at point B.
The last column of Table 2.4 shows the product
Example 2.18 Application indeterminate problem
Table 2.4 Calculation of horizontal movement of th e roller support B support B
Force L/AE −8 −1 kN (×10 N )
Fi,
method
to
a
statically
The problem is quite easily solved by making the point B free to move horizontally but applying a horizontal force Q ( in kN) at this point to Fig. 2.48 Replacing reaction with an undetermined force control its movement. The strategy consists of calculating the horizontal movement of point B in terms of the load Q, and then determining the value of Q for which this movement movement is zero. It should be clear that that this value of Q for which the movement is zero is the value of the horizontal reaction at point B.
−4
0 1 0 1 0 0 0 0 0 Total
energy
Solution:
(×10 m)
1.24 1.59 1.24 1.59 1.24 1.24 1.59 1.24 1.24
of
Let us once again consider the same truss as in Examples 2.16 and 2.17 and shown in Fig. 2.42 with one change: The roller support is replaced by a pinned support. This is as shown shown in Fig. 2.47. This will introduce a horizontal reaction at support point B. There will now be four reactions and only three equilibrium equations to Fig. 2.47 A statically indeterminate truss determine them. The problem, therefore, is statically indeterminate. indeterminate. We illustrate below the general strategy to solve solve such problems using energy method.
Table 2.4 organizes the calculation of the horizontal movement at the roller support at B. Here the values in the second and the third columns ( F i, and L/AE ) are the same as in Table 2.3, but column 4 shows the values of the axial forces in the various members when only the load Q (equal to 1 N) is applied at point B. As was noted above, only two non-zero entries (for members AF and BF ) result.
−16.7 AD AF + 13.3 BE −25.0 BF + 20.0 −16.7 CD CE −16.7 CF +5.0 DF 0 −8.3 EF
for each member of
the truss. The sum of these over all the members members gives the horizontal deflection deflection − of point B (in-line with the dummy force Q), which is seen as 52.9×10 5 m, slightly more than half a millimetre.
Here again, we use the same procedure Fig. 2.46 FBD for a unit horizontal load at B at B as above. The calculation of F i’s is does not depend on the dummy loads and we get the same values as in column 2 L/AE for the various members are also unchanged. of Table 2.3. The values of L/AE We, however, have to calculate dF i’s under the unit lod Q. For this purpose we we use an FBD as shown in Fig. 2.46 and calculate the resulting tensions in each of the members. members. Clearly, the the vertical reactions R1y and R2y are zero, and the horizontal reaction R1x is – 1 – 1 N. It is easy to see that there will be tension only in the members AF and BF , both +1 N.
Member
0 21.1 0.00 31.8 0 0 0 0 0 52.9 37
, where δV is the volume of the the element. For a finite body
The vertical components of the reactions remain unchanged from the determination in Example 2.16: and . The forces forces in the various members can be calculated by considering the FBDs of the various pins. This is quite straight forward, forward, and the values so obtained are tabulated as column 2 in Table 2.5 below. The values in column 3 and 4 of this table will be
then we can find the strain energy by integration over the whole volume:
∫
Table 2.5 Calculation of horizontal movement of the roller support B support B
Member
L/AE Force −8 −1 kN (×10 N )
Fi,
It stands to reason that the stress ζ xx does not do any work with the strains ε yy and ε zz, since these strains result in displacements which are perpendicular to this stress. Similar work will be done by the other stress components.
−5
(×10 m)
−16.7 AD AF + 13.3 + Q −25.0 BE BF + 20.0 + Q −16.7 CD −16.7 CE CF +5.0 DF 0 −8.3 EF
1.24 1.59 1.24 1.59 1.24 1.24 1.59 1.24 1.24
0 1 0 1 0 0 0 0 0 Total
`
(2.15)
0 21.1 + 1.59 Q 0.00 31.8 + 1.59 Q 0 0 0 0 0 52.9 + 3.18Q
Let us now look at the work done by shear stress components. Consider a shear element as shown in Fig. 2.50 where the lower surface is anchored and the upper surface has a shear stress of η of η yx. Let the element undergo undergo a shear strain of γ yx as shown. shown. This implies that the upper surface with a force of η of η yxδ xδ z acting on it undergoes a displacement of γ yxδy. δy. The strain energy of this infinitesimal element, thus, is
the same as in the corresponding column of Table 2.5, and then column 5 is modified as shown. The horizontal displacement displacement of point B under the action of a − force Q at that point is thus (52.9 +3.18Q) ×10 5 m, with Q in kN.
()( )
This displacement is zero when Q = − 52.9/3.18 = − 16.6 kN. Since the ctul pin B is restrained from moving, the reaction at point B must be 16.6 kN, inwards.
Fig.2.49 An infinitesimal element under tension
Fig. 2.50 A Shear element
For a finite body then we can find the strain energy by integration over the whole volume:
We have, thus, solved the statically statically indeterminate problem. Once this is known, we can calculate loads in all the members of the truss.
∫
`
(2.16)
Similarly for the other shear components. components. Combining the strain strain energies for normal and shear strains, we can write the equation for the general state of strain for a body of volume V as
2.11 Strain energy in an elastic body We have so far considered the strain energy of axially loaded structures. structures. Let us extend this to a more general state of stress and strain. Consider an infinitesimal δz along the three cubical element of dimensions δx, δy, and δz along three coordinate axes. Let this be acted upon by a force which produces a stress ζ xx in the x-direction. The force acting on these faces is . If the strain in the the x-direction is ε xx, the elongation is ε xxδx. The strain energy of this infinitesimal element, thus, is
∫ ( ) `
38
(2.17)
Summary
The general strategy for solving the problems of the mechanics of deformable bodies consists of three major steps: 1. Consideration of static equilibrium and determination of loads in members of the structures. This equilibrium analysis invariably requires drawing up strategically chosen free bod y diagrams (FBDs), 2. consideration of relations between loads and deformations, and 3. consideration of the conditions of geometric compatibility.
where ν is termed as the Poisson ratio and is a property of the material. With this, the net longitudinal strain in any direction is related to all the three tensile stress components.
( ) ( )
The second step above requires the considerations of the materials as well as the geometry of the structures. This step in itself may be divided into into three distinct phases: a. b. c.
(2.9)
, and
We convert loads to stresses in the various members (macro to micro ), the stresses are then converted to strains using the material properties (micro to micro transformation ) , and finally, with the use of the geometry of the structure, we determine deformations from the strains so calculated ( micro to macro ).
We define the shear strain γ xy in the element at a point as the distortion in the shape of the element as measured by the change in angle between two mutually perpendicular lines (parallel to x- and y- directions).
The shear stress is related to shear stress by the relation: . Shear strain in a plane depends only on the shear stress in that plane and no other stress component. Heating of a material material too does not introduce any shear strain.
The strength of this method lies in the fact that the first and the third steps above require the consideration of the geometry of the structure (and are independent of the material), while the second step requires considerations of only the material of the structure. structure. Thus, the stress-strain relation obtained from the tensile test with its most simple loading conditions can be used to predict behaviour of very complicated structure, since the second step above is independent of loading and geometry and depends entirely on the material.
Change in temperature of a structural member introduces thermal strains so that the net strains of an element subject to a general state of stress and to a change in temperature is given by:
We may, at time, need to go in the reverse direction. direction. We may be given the total deformation from which we need to determine the loading: we calculate strains from deformation, convert strains to stress using material properties, and then integrate stress stress to determine the loading. The strategy here too is macro to micro to micro to macro. There are many problems in which we cannot take the steps enumerated above in a linear sequence, because there are not enough equations of equilibrium to solve for all the unknown loads. In such cases we have to consider simultaneously all the three steps, even if we were interested in only one, say, in determining the forces in the system. Such problems are are known as statically indeterminate problems. Examples 2.3 to 2.5 illustrate the methodology to be adopted for solving such problems. As we apply an axial load to a member and it develops a strain ε xx in the axial direction, there is also a transverse strain in the y- and z- directions. The strain in a transverse direction is a fixed (negative) proportion of the strain i n the axial direction.
39
The stress-strain properties of a material are determined by a tensile test. test. In Sec. 2.6 we introduced the following concepts: Proportional limit : The stress up to which the stress is linearly proportional to strain. Elastic limit : The stress up to which the material upon unloading returns to its undeformed length. Both proportional and elastic limits limits are difficult to determine. Permanent set : The residual strain in a material after unloading. Yield strength : The stress level at which the permanent set is 0.2% (or the residual strain is 0.002). A material is modelled modelled as elastic up till this level of loading. Strain hardening refers to the increase in yield strength of a material due to its plastic straining. It is exhibited by materials in which the stressstrain curve slopes upward in some part of the plastic region. Ultimate strength is the maximum nominal stress in a material before necking begins. Necking refers to sudden and localized sharp reduction of cross-sectional area. Most of the differences between the behaviour of ductile and brittle material arises from the fact that there is little plastic deformation in a brittle
material. A brittle material undergoes fracture with very little strain, much less than that for for a ductile material. Glass, for example shows no plastic deformation at all. Another feature to note is that the strength to fracture is much larger in compression than in tension. As a structure is loaded the displacements of the points of application of external loads result in work work being done. This work done is stored as as the strain energy of the body given as
∑
bar by an axial load is calculated as
.
Castigliano theorem states that if the total strain energy of a structural system is expressed in terms of the external loads applied to it, the deflection at any one loading point (in-line with the load) is obtained by taking the partial derivative of the strain energy with respect to the load at that point:
. The work done on a simple
.
The procedure consists of applying a load P at the point where the deflection is to be calculated and in the direction in which the deflection is desired. This may involve replacing an existing load with P, or inserting a dummy load. The loads F i’s in the vrious members re clculte in terms of the unknown load P, and then the total energy U is calculated. After taking the partial derivative of U with respect to P, the actual value of P is plugged in (or P is replaced by zero, in case it is a dummy load). This gives the desired deflection. In the case of more involved trusses, it was shown that we could simplify the calculation of the deflection at the location of a load P by rewriting the partial derivative as:
∑ ∑ ∑
.
Here we first calculate F i’s without the lod P, and then calculate the value of as the increment in ’s for unit lod P. This is known as the unit load method whose calculations can be organized as in Table 2.3 or 2.4 Statically indeterminate problems can also be solved using the energy method. The strategy consists of relaxing one of the boundary constraints by letting that boundary point move. A dummy load is introduced at that location in place of the reaction that would have been there if the constraint was not relaxed. The deflection of the boundary point is then calculated in terms of that dummy load. We next obtain the value of the dummy dummy load required to make the movement movement of that point as zero. This value is the desired reaction at that point. The strain energy per unit volume of the body is given b y
( ) The total strain energy can be obtained by integrating it over the entire
volume of the body.
40
To obtain the relations between twisting moment, angle of twist, and the stresses and strains we shall follow the strategy that we outlined in the last chapter, but in a reversed order. The reason for this is that unlike the situations situations involving axial loading treated in the last chapter, it is not possible to make the assumption of uniform stresses in cases cases of torsional loading. In fact, it will be shown that the stresses on the cross-section of a shaft vary with the distance from the axis of the shaft. For a constant diameter shaft, we shall first assume a total angle of twist θ for an applied twisting moment T . We then use the geometric geometric considerations to convert convert this macro quantity to the strain at various various points in the shaft. This is the macro to micro stage. We next use the material properties to convert strains to stresses. This is the micro to micro transformation. In the last stage we convert convert the stresses into torque, a micro to macro conversion. This completes the solution to the problem.
3 Torsion of circular shafts
3.2 Relating angle of twist to twisting moment
3.1 Introduction
Fig. 3.2 shows a circular shaft25 of radius R and length L clamped at one end and subjected to a twisting moment T . Let the shaft undergo a twist through an angle θ as shown. We first determine the strain components in the shaft. For this purpose we make some simplifying assumptions26:
A slender rod which is subjected primarily to twisting or torsional load is termed as a shaft. One of the more important important uses of torsional members is in the transmission of power or motion through circular shafts. A shaft is used to transmit the torque generated by an electric motor to the load it rotates. The torque produced by the engine of a car is transmitted to its wheels by a shaft. When a coil spring elongates, the wire constituting its coil is Fig. 3.1 Torsion bar used as suspension in automobiles twisted. Torsion bars are also used as springs in automobile suspension (Fig. 3.1).
Fig. 3.2 A circular shaft under torsion
Circular cross-sections of the shaft perpendicular to the axis remain plane, i.e., there is no warping of the shaft, cross-sections of the shaft do not deform, i.e., there are no strains within a cross-section of the shaft. Radial lines within the cross-sections remain straight and radial, and there is no change in length of the shaft, i. e., there is no axial strain.
In problems involving torsion of shaft we may be interested in determining:
The stresses that result when the shaft carries a specified twisting moment. We may consequently determine determine the maximum twisting twisting moment that the shaft can carry without failure. The twist in the shaft when carrying a specified twisting moment. moment. In application to torsional springs, we may like to determine the spring constant, which is the angular twist per unit twisting moment.
25
We assume a circular shaft because many of the assumptions made here a re not valid for non-circular shafts 26 These assumptions can be shown to be true by using some fascinating symmetry arguments. arguments. See Crandall, Dahl Dahl and Lardner, An introduction to the mechanics of solids, 2 ed., McGraw-Hill
41
The above assumptions imply that there are no linear strains. Therefore, using the cylindrical polar coordinates as shown,
Conversion of strain to stress is simple. simple. Since there is no strain other than the the shear strain γz (and its complementary shear strain γ z ), there is only one component of strain η z z (and its complementary shear stress η z ). These stress components on an element are shown in Fig. 3.5. The shear stress component can be obtained by multiplying the strain with the shear modulus G:
No distortion of the sections rules out shear strains and γr and γ zr also vanish. The only non-zero component of strain is γr which we now proceed to determine.
Calculation of strain from macro distortion
For this purpose we take a slice of length dz at a distance z from the clamped end as shown in Fig. 3.3. The detailed geometry of distortion is shown in Fig. 3.4. Let the twist of the face at z be θ, and that at z + dz be θ + dθ. dθ. To calculate the shear strain in the shaft, consider a ‘rectngulr’ element ABCD on the surface of the cylinder. After distortion, the shape of this element changes to A1 B1CD. The angular twist dθ
(3.2)
Converting stress into loading Fig. 3.6 The shear stress distribution on Consider an elemental area in the the cross-section of the shaft cross-section of the shaft at radius r and angle as shown in Fig. 3.7. The shear stress stress at this location location is
. If the area of the element is dA, the shear force on this element is . It is easy to see that the force on an element of equal area which is diametrically opposite to this element is equal in magnitude but has the opposite
Fig. 3.3 The slice whose distortion is studied
direction. This is true of all elements of the crosssection and, therefore, the net shear force on the section is zero. But that is not true for the moment moment about the origin. origin. Every elemental elemental shear force contributes a counter-clockwise (hence, positive27) moment, so they add up. up. The sum of moment can be found by simple integration. integration. The force on the element shown at a radius of r is stress times its area,
the strain at r = R is .
or,
Fig. 3.4
. Therefore, its contribution to the
twisting moment about O is radius times this force,
It can be seen that the strain γz is not constant with r . Consider the deformation deformation at a radius of r . Here, the the point E has dz, or, shifted to point E 1, so that strain at this radius is given by EE 1 / dz .
Fig. 3.5 Shear stresses on an element
where G is the shear modulus of the material. Please note that that neither neither the geometry nor the nature of loading enters this step. The variation of shear stress with the radius r is shown in Fig. 3.6.
of this element is the angle that the movement from A to A1 subtends at the axis passing through point O on the axis of the shaft. What, then, is the strain? We see that which was originally a right angle has now changed to . Thus, it has reduced . Therefore, the strain here is equal by DA. Thus, to which is seen as AA1 /
,
27
Fig.3.7 Shear force on an elemental area
Consistency of results requires that we define our co-ordinate directions and sign conventions very clearly. clearly. It is common to use only rightright- handed triad for the direction of axes: Curl the fingers of your right hand from the positive x- to the positive y-axis. The thumb then points in the positive positive z-direction. We shall use only right-handed triads in this book.
(3.1)
Converting strain to stress
42
or,
∫ ∫ ∫
The maximum shear stress in a shaft occurs where the value of r is maximum, I zz from Eq. 3.6 in Eq. 3.4, we get i.e., at r = R. Using Value of I
. Thus, the total total twisting moment moment T is
The integral is termed as the second moment of the area about the polar axis of the section. Many authors term this as the polar moment of area. It is a geometric property of the area and is denoted by28 I zz. This reduces the expression for the twisting moment T to
Thus, the maximum shear stress in a shaft for a given twisting moment T 3 decreases as D . We illustrate below the use of these formulae.
(3.3)
Example 3.1 Diesel generating set The shaft of a 2 MW diesel generating set set rotates at 500 RPM. What is the minimum diameter of the steel shaft connecting the diesel engine to the electric generator if the shear stress in not to exceed 40 MPa?
We have completed the solution of the problem of the torsion of a shaft under a twisting moment. We can recast Eqs. 3.1-3.3 to obtain any quantity of interest. dθ/dz from Eq. 3.3 in the expression for the shear stress η, Replacing dθ/dz from
Solution:
(3.4)
We first calculate the torque from the po wer being transmitted: P transmitted: P = T×ω. Here ω 6 is the angular speed equal to 2π× 2π×RPM/60 RPM/60 = 52.36 rad/s. Torque T then is (2×10 4 W)/(52.36 rad/s) = 3.82×10 N.m. This is the twisting moment in the shaft. shaft. The shear stress varies across the section of the shaft with the maximum stress occurring at the surface at r = R, where R is the radius of the shaft.
The total twist θ is obtained by integrating dθ/dz over the entire length of the shaft:
∫ ∫
(3.5)
To calculate the stress in the shaft at r = R, we use Eq. 3.4 with the value of I zz given by Eq. 3.6:
The product GI zz is termed as the torsional rigidity of a shaft. The more its value, the less is the twist produced for a given length and t wisting moment
3.3 Stresses and strain in a circular shaft
4 4 With T = 3.82×10 N.m and η ( R) = 40 MPa, we get R = [2×3.82×10 N.m/(π× N.m/(π×40 40 1/3 MPa)] = 0.085 m, or 8.5 cm. Thus, the diameter of the the shaft needs to be at least 17 cm.
We now evaluate the value of I zz for a circular shaft of radius R. We replace the elemental area dA of Fig. 3.7 with its value rddr, and integrate over from from 0 to 2π , and over r from 0 to R:
∫ ∫ ∫ ∫ ∫ for a circular shaft.
What is the twist per meter run of the pipe?
, or
To obtain the twist angle θ per meter run, use Eq. 3.5 with L = 1 m:
) ( , -
(3.6)
The torsional rigidity GI zz of a circular shaft varies as the fourth power of its radius or diameter.
, or 0.133o per meter run,
a negligible amount.
Example 3.2 Torsion bar Fig. 3.8 shows a torsion bar used as suspension in an automobile. The load from the wheels is applied to a rigid arm as shown. Calculate the spring constant for the bar.
28
Many texts use the symbol J to denote the second moment of area about the z-) axis. We prefer here the symbol I with two subscripts denoting the polar ( zdistance from the specified axis.
43
the right-hand thumb29 rule for directions of torques, it is pointed in the positive z-direction, and hence, positive. We arbitrarily show the twisting moment moment at z to be positive T . The moment balance of this FBD FBD shows that T must be −40 kN.m. z between 0 and 1.5 m. This is valid for all vales of z
Solution: The twist in the bar then is given by Eq. 3.5 as:
. For
z is larger than 1.5 m, the resulting FBD is as shown in If, however, the value of z Fig. 3.10b, and the value of T is −20 kN.m. The vritions of T with the value of z have been plotted in Fig. 3.10c. This is known as the Twisting or Torsion 30 Moment Diagram (TMD)
an applied force F, the torque T is F×(0.7 m), L = 2 m, G for steel is 80 GPa, and I zz = Fig. 3.8 Torsion bar (0.04)4 /32. Plugging π (0.04) in the values, we get −5 for a load F , a twist of 6.96×10 F . This twist results in an upward motion of the load point of 0.7×θ, 0.7×θ, or 4.87×10−5F . Thus, the effective effective spring constant for for the torsion bar spring is k = 20.53 kN/m.
The value of shear stress is given by Eq. Eq. 3.4. The maximum shear stress will occur in the section AB where the twisting moment is the maximum. And it will be at the surface where the value of r is the maximum (equal to R):
Example 3.3 Power shaft Fig. 3.9 shows a power shaft of diameter 4 cm transmitting power to two machines. The shaft is driven by a belt drive that applies a torque torque of 40 kN.m at pulley A. The power take-off is again through belts at pulleys B and C. if the shaft diameter is 8 cm, determine the total twist of the shaft and the maximum value of shear stress in the shaft.
[The diameter of the shaft is 8 I zz cm throughout. throughout. The value of I is πD4 /32, or π (0.08 (0.08 m)4 /32 = −6 4 4.02×10 m .]
Fig. 3.10 Twisting moment diagram
This is a fairly large value, close to the failure strength of structural steel. We probably need a thicker shaft shaft to give it a margin of safety. The twist of the shaft section AB of length L = 1.5 m which carries a uniform T of −40 kN.m is given by Eq. 3.5 s Fig. 3.9 Power shaft
Solution:
29
Right hand thumb rule states that if you curl the fingers of your right hand in the sense of the torque, the thumb points in the direction of the moment vector. 30 These TMDs are a big help in the analysis of shafts which do not have uniform twisting moments along them. them. We can draw the TMDs directly without without drawing FBDs if we follow the following procedure: (1) start from T = 0 at the left most end; (2) at every location where a concentrated moment T o acts, decrease the value of T by T o, if T o is positive, or increase by T o if it is negative. We will come across similar procedures for drawing shear force diagrams (SFDs) and bending moment diagrams (BMDs) in Chapter 4.
We first draw a diagram depicting the variation of twisting moment along the axis of the shaft. To determine the twisting moment T at any location in the shaft, we externalize the moment there by taking a section at that point and draw an FBD. Fig. 3.10a shows an FBD for determining the value of T for all values z between 0 and 1.5 m. of z m. The torque acting on pulley A is 40 kN.m, and using
44
∫
.
I zz in terms of the diameter D( z), which, in turn, is We insert here the value of I z as above. expressed in terms of z
A reasonable value of G for steel is 80 GPa. Similarly, the twist of shaft section BC is exactly half of this, since it carries only hlf the twisting moment (−20 kN.m), i.e., 0.093 rd. The totl twist of the shft is the lgebric sum of the two: −0.279 rd or −16 o. The negative sign indicates that the end at C twists clockwise with respect to the end A.
degrees.
∫ ,* ,* + +-
= 0.024 rad or about 1.4
Where and what will be the magnitude of the maximum shear stress in this tapered shaft? The shear stress is maximum (at a given given section) at the outer most location, i.e., at r = R. Thus,
Example 3.4 Twist in a tapered shaft
We next consider a tapered shaft in which the polar moment of area, and therefore, contribution to the twist varies along the axis of the shaft. Fig. 3.11 shows a 1 m long tapered steel shaft subjected to a twisting moment of 1 kN.m. If the base diameter is 10 cm and the tip diameter is 7 cm, determine the Fig. 3.11 Tapered shaft total twist of the shaft.
4 Since I zz varies as R , and T is constant along the shaft, the value of maximum η along the length of the shaft varies like R−3, i.e., the maximum value of the shear stress along the length occurs where R has the minimum minimum value. The maximum π×(0.07 m)4 /32 = shear stress, therefore, occurs at the tip where D is 0.07 m, I zz is π×(0.07 −6 4 2.36×10 m , and T = 1 kN.m.
Solution:
Example 3.5 Cable in a sleeve
The contribution dθ to the twist of the shaft by a section of length dz at z is given by (T/GI zz)dz. In this this problem the twisting moment T is constant along the length of the shaft, but I zz varies with z. It can be verified that the linear variation of diameter D is given by . The value of I zz varies as fourth power of the diameter, i.e., as Fig. 3.12 Variations of twisting πD4 /32. Fig. 3.12 shows the moment, area, polar moment of variations of the twisting area, and the contribution to twist moment, area, the polar moment along a tapered shaft of area, and the contribution to twist along the length of the tapered shaft. The total twist can be obtained by integrating the contribution of an elemental elemental length. Thus,
Fig. 3.13 shows a 4 mm dia long cable in a sleeve used to control a setting remotely. As we turn turn the knob the connecting cable twists inside the sleeve. The sleeve is provided so that the cable does not flip. The friction between the sleeve and the cable applies a torque on the cable which needs to be overcome. This torque is estimated to be about 0.3 Fig.3.13 Cable in a sleeve N.m per meter run of the cable. If the torque needed at the setting point is 0.5 N.m, determine the maximum length of the cable, if the cable material can withstand a shear stress of 100 MPa. What would be the play? (The play play is the angle through which you need to turn to effect a change in setting when you reverse the setting.)
Solution: Fig. 3.14 shows the TMD of a cable of length z. The maximum torque in the cable is (− (− 0.5 −0.3 z) N.m, and occurs at the knob-end. The maximum shear shear
∫ 45
stress would be at the surface of the cable, i.e., at r = 0.002 m. m. Eq. 3.4 gives
kN.m. This torque torque is negative, i.e., i.e., clockwise looking from the right end. Fig. 3.16 shows the twisting moment diagram for the two shafts.
m
N.m N.m
m
Using the condition that maximum shear stress is limited to 100 MPa31, we get the maximum value of z of z as [(100×106 /79.6×106) – 0.5]/0.3 m, or 2.5 m.
The maximum torque in a shaft is given by , where the radius of the the shaft. The R is the maximum value of the shear stress in the first shaft is
The maximum twisting moment, then, is (−) 1.25 N.m, and the twist is given by Eq. 3.5 as
Fig. 3.16 Twisting moment diagram for . This This is quite safe. The geared shafts maximum value of the shear stress in the second shaft is . Thus, the maximum (shear) stress in the shafts is 70.8 MPa.
Fig. 3.14 TMD of the cable
, - m
, - , , -
or 88o. This is the play in in the cable.
Let us next find the total twist at the free end. The twist in the each of the two shafts is obtained from the equation .
Example 3.6 Geared shafts
Fig. 3.15 show two steel shafts geared together. One end of the first shaft is built in a wall. Find the maximum shear stress stress in the shafts and the total rotation of the free end of the second shaft.
, - - Twist
in
the
first
shaft,
,
counter-clockwise
looking from right.
Solution:
Twist in the second shaft (with respect to the geared end of the shaft,
The first step in solving this problem is to determine the twisting moments along the two shafts and to draw the twisting moment diagrams. Clearly, Fig. 3.15 Geared shafts the torque in the thinner (second) shaft is 3 kN.m throughout its length. To determine the torque in the thicker (first) shaft one one has to be careful. The torques in the two shafts are not equal. (Why? Why does a simple torque balance not give the correct result? 32) The problem can be solved if it is realized that the contact force F between the two gears must be equal and opposite. And therefore, the torque in in the first shaft is 10/6 of 3 kN.m, or 5
, , - , clockwise looking from right.
To find the total twist of the free end with respect to the fixed end of the first shaft we cannot just add the two twists. Consider the motion of the gears gears as shown in Fig. 3.17. Angle 1 represents the clockwise movement of the Fig. 3.17 larger gear (on shaft 1). It is clear that angle 2 which Geometry of twist represents the counter-clockwise movement of the smaller gear due to gearing alone is 10/6 of of 1. This is the rotation of the left end of shaft 2. Since 1 is 0.0064 rad, angle 2 is 0.01 rad counter-clockwise. counter-clockwise. The twist of shaft 2, which is 0.03 rad counter-clockwise is superimposed on the motion of the gear, to obtain the total rotation of the free end as 0.01 rad + 0.03 rad = 0.04 rad or 2.3 degree.
3.4 Hollow shaft 31
We shall see here that hollow shafts provide excellent rigidity for weight of material. Consider a hollow circular circular shaft of inner radius Ri and an outer radius of R Ro as shown in Fig. 3.18. The value of I zz can be obtained by replacing dA in r 2dA with rddr and 2π and and over r from Ri to Ro. rddr and integrating over from from 0 to 2π
We need to use a negative sign with η max max here. It is because we have not shown the bearings that will be required to hold the shafts in place. The bearings will apply forces on the shaft which will produce moments that should also go in the moment balance equation. 32
46
∫ ∫ ∫
cannot be determined by considerations of equilibrium equilibrium alone. In such shafts the geometrical constraints on twists need to be invoked to solve the problems. We follow the same three-step process that was invoked in the last chapter:
(3.7)
The maximum shear stress in the shaft will b e near the surface, i.e., at r =Ro. Therefore, the maximum stress in the shaft varies inversely as I zz. If we define the torque carrying capacity of the shaft as the maximum torque the shaft can carry without the stress exceeding a Fig. 3.18 Hollow specified limit, it is clear that this varies directly as cylindrical shaft I zz (for a fixed value of R Ro). The curve the value of I labelled (a) in Fig. 3.19 which plots the variation of I zz (as the fraction of the I zz of the solid shaft of radius Ro) also represents the variation of the torque carrying capacity. The curve (b) plots the variation of the weight (which varies as the area of cross-section) of the shaft. The curve (c) shows the torque carrying capacity per unit weight of the shaft showing clearly the cost effectiveness33 34 of a hollow shaft .
⁄
3.5 Statically indeterminate shafts There are many situations in which the twisting moments in shafts
Use equilibrium analysis to write equations for the twisting moments in the various parts of the shaft. There may not be enough equations to determine the moments explicitly. Convert twisting moments to twists in each part. Write geometrical compatibility conditions to complete the analysis.
We illustrate the procedure with a few examples.
Example 3.7 A shaft built in at both ends Consider a composite steel shaft ABC built up of two 1 m long steel sections, AB of 10 cm dia and BC of 15 cm dia (Fig. 3.20). A torque of 100 kN.m is applied to the collar in the middle. The two ends of the shaft are built into walls so that there is no rotation of the ends. What is the twist produced in the shaft?
Solution: Equilibrium analysis (a) (b) (c)
Variation of the value of I I zz of hollow shaft Variation of the weight of hollow shaft Variation of the ratio of strength to weight
Let the reactions at the two ends be T 1 and T 2 as shown. From the condition of the equilibrium of moments we get T 1 + T 2 + 100 kN.m = 0. It should be clear that either T 1 (a) The composite built in shaft or T 2 or both should have a (b) FBD for determining torques in part AB part AB negative sign. This is the only (c) FBD for determining torques in part BC part BC equation we get from the (d) Torsion or twisting moment diagram statics for the two unknown Fig. 3.20 reactions T 1 and T 2. Clearly this is a statically indeterminate problem. We need to consider deformations as well.
Fig. 3.19 Comparison of hollow shaft with solid shaft for various ratios of the internal and external radii
33
This, of course, has not taken into account the increased manufacturing cost of the hollow shaft.
34
This is quite as expected. An element nearer the surface has larger stresses and contributes more to the torsional moment. Therefore, the material nearer the surface is used more effectively to bear the torsional load than an equal area near the axis. Seen in another way, it is an area nearer the outer surface that makes a larger contribution to I zz, the value of the polar moment of inertia.
47
−
θ AB = −1.92×10 6T 1 = −6 (−1.92×10 ) × (−16.6 kN) = 0.032 rad
We need to determine the twisting moments in the two halves to calculate the twists therein. For this purpose, we take a section first in the part AB and draw an FBD for the left part as shown in Fig. Fig. 3.20b. Here T is the twisting moment in the shaft. The equilibrium of this FBD gives T 1 + T = 0, or T = − T 1. Fig. 3.20c shows the FBD for determining the twisting moment in the second part of the shaft. From this we get T 1 + T + 100 kN.m = 0, or T = − T 1 – 100 – 100 kN.m. Fig. 3.20d shows the complete twisting or torsion moment diagram (TMD) of the 35 shaft . If we have chosen the sign of T 1 correctly, then both halves of the shaft have negative twisting moments36.
Example 3.8 Meshed gears Fig. 3.21 shows two shafts, one of dia 6 cm and the other of 4 cm, carrying identical meshed gears in the middle. The length of each shaft is 1 m. The two shafts are disengaged, the top o shaft A rotated by 10 in the direction shown, and then the two gears are meshed again. Shaft A will untwist a bit twisting shaft B till the equilibrium is reached. Determine the locked in torque at equilibrium.
Relating torques and twists
Now that we know the twisting moments, we can calculate the twist using Eq. 3.5: .
⁄
The values of I I zz are: for shaft AB: πD4/32 = π(0.1 m)4 /32 = 9.81×10−6 m4 4 4 −6 4 for shaft BC : πD /32 = π(0.15 m) /32 = 49.7×10 m
Solution:
, and
Fig. 3.22 TMDs of the two shafts under the assumptions that the reactions on the left wall are T 1 and T 2, respectively.
Equilibrium requires that the contact force between the gears be equal and opposite. Since the two gears are identical, the torques in the two shafts are equal but opposite. Let this residual torque be T o, positive T o in (the left part of) shaft A and negative T o in (the left part of) shaft B.
Geometric compatibility
Since both ends of the shaft are built in, the net twist in the shaft must be zero: θ AB +θ BC =0. This gives:
Let us assume that wall at left applies on shaft A a reaction torque T 1, and on shaft B a reaction torque T 2. The TMDs of the two two shafts can be calculated calculated and are as shown in Fig. 3.22. These have been drawn under the assumption that T 1 and T 2 are both positive.
−1.92×10−6T 1 – 0.38×10−6T 1 – 0.038 = 0, which gives
Now let us look at the deformation of shaft A. It is clear that the twist in the left half must be the same but opposite of that in the right half, for the total twist from end to end to be zero. zero. Since the two halves have identical geometries, it stands to reason that the torque in the two halves must be equal and opposite. There is, thus, thus, only one solution: the twisting moment in the left half is + T o /2, and that in the right hlf is − T o /2. Similarly, we can argue that the twisting moment in the left half of shaft B be − T o /2, and that in the
T 1 = −16.6 kN
With this value of T 1, we can calculate the twist at point B as
(a) TMD for top shaft (b) TMD for bottom shaft
Fig. 3.21 Two meshed shafts
35
The reader can verify that we could have drawn this TMD directly without drawing FBDs if we had followed the procedure outlined in footnote 5. 36 It is clear that the sign of T 1 that we have chosen here cannot be correct because with this we have negative twisting moments in both halves of the shaft. This, of course, is not possible since if one part twists clockwise, the other must twist counter-clockwise. But it does not matter for our analysis here.
48
(a) TMD for top shaft (b) TMD for bottom shaft Fig. 3.23 Actual TMDs of the two shafts
right half be + T o /2. These are shown in Fig. 3.23.
strain with the value of G = G 1 for the core material, and in the outer sleeve for r between r 1 and r 2, we multiply with the value of G = G2 for the sleeve material, which is higher. Thus, the graph for shear shear stress is kinky with two two different slopes, one for the the core and the other for the sleeve. This is shown in Fig. 3.26.
Geometric compatibility
Refer to Fig. Fig. 3.24. Shaft A was rotated by 10o when the two gears were were engaged. After engagement and letting go, the shaft A unwinds by an angle θ2 which is the same as winding up of shaft B (since the gear ratio is one). one). The residual twist in shaft A is now θ1. It should be clear that that the geometric geometric compatibility condition is:
The next step in Section 3.2 was considering the shear force on a small element, calculating its contribution to the twisting moment about the axis, and integrating over the entire cross-section:
Fig.
3.24
∫ ∫
Geometric
compatibility condition
θ1 + θ2 = 100 = 0.175 rad
Here, unlike what we did in Section 3.2, we cannot take G outside the integral since it is not constant over the section. But we can do it part-wise: we divide the area in to two parts, one, with r between 0 and r 1, which covers the core with modulus G1, and the other with r between r 1 and r 2, which covers the sleeve with modulus G2.
We can calculate the twists θ1 and θ2 using Eq. 3.5: θ = TL/GI zz. Thus,
, , - , , - m
G
m m
G
m
, and .
Using these in the geometric compatibility condition, we get 2.46×10−6T o + 12.45×10−6T o = 0.175 This gives T o, the locked-in twisting moment as 11.7 kN.m.
3.6 Composite shaft Let us next consider twisting of a composite shaft that is made up of inner core of one material clad with an outer sleeve of another material as shown in Fig. 3.25. To solve this problem, let us look carefully through the derivation of the torsion formulae in Section 3.2 and figure out the changes we would need to make on account of two different materials in the cross-section.
or
(3.8)
(3.9)
The torsional rigidity of the composite shaft is, thus,
. This does not change change with
.
the material. material. Thus, here here too, stain γ z varies linearly with the distance r from the longitudinal axis, independent of material. We had next converted the shear strain to shear stress by multiplying it with G, the shear modulus. Here the material material dependence arises. arises. In the inner core up to r = r 1, we multiply the
∫ ∫ ∫ 2∫ ∫ 3 2 ∫ ∫ 3
where I 1 is the polar moment of the core area and I 2 is the polar moment of the sleeve area. Compare this with Eq. 3.3. The term GI zz for a uniform shaft is replaced with (G1 I 1 + G2 I 2). Similarly,
We had assumed a twist angle θ and showed using geometry that there is only one shear strain component γ z which is given by Eq. 3.1 as
Fig. 3.26 Shear strain and shear stress distribution across a composite shaft
Here
Fig. 3. 25 A composite shaft
49
, and
.
shear force on this element is qδs whose moment about an arbitrary point O is hqδs, hqδs, which is 2qδA, 2qδA, where δA is the grey area in Fig. 3.28a. The total twisting moment, then, is simply the shear flow q times twice the area A of the cross-
3.7 Torsion of thin-walled tubes We have so far found exact solution to the problem of torsion in circular (and hollow circular) shafts. shafts. It is not possible to find such solutions solutions for shafts of arbitrary cross-sections. cross-sections. However, we can find quite easily the approximate solution for a thin-walled shaft of arbitrary section. Consider such a shaft as shown in Fig. 3.27. We use the n, s, z co-ordinate system as shown, with the coordinate axis s along the central line of the tube wall. The dominant stress component in this case is η zs (along with the complementary stress component η sz We shall assume, sz). without proof, that the other stress components are either absent or are negligibly small.
(a)
Fig. 3.28 Relating shear flow to the twisting moment
section of the tube: flow38 as:
We show in Fig. 3.27b a small (a) (b) element of length dz of the shaft. The shear stress η zs is distributed over Fig. 3.27 A thin-walled torsion tube the top and bottom surface of the element while the complementary shear stress η sz sz is distributed over the surfaces marked 1 and 2 in the figure. We introduce here the concept of shear flow q defined as the shear force per unit length of the tubular surface:
∫
(b)
, from which we can extract the value of the shear
(3.11)
Example 3.9: A hollow circular shaft Let us evaluate the stresses in a hollow circular shaft by the approximate method outlined in Sec.3.7 and compare the results with those obtained by the exact method. Consider the hollow circular circular steel shaft of length 1 m shown shown in Fig. 3.29. It is subjected to a twisting moment T = 100 N.m. Compare the stresses stresses obtained by the exact and the approximate methods.
(3.10)
In Fig. 3.27b we have shown the net shear forces on the vertical surfaces as shear flow q times the length dz of the surface. The first thing to notice is that shear force per unit length on the vertical surface 1 is the same as on the horizontal surface. The next thing to notice is that the two forces shown shown are the only significant vertical forces on the element, and therefore, q1 = q2, that is, the shear flow is the same at two arbitrary positions, and hence everywhere, along a crosssection of the tubular shaft37.
Solution: Eq. 3.4 gives the stresses in a circular shaft:
, where I zz for a hollow
shaft is given by Eq. 3.7 as
.
We can now relate the shear flow q to the twisting moment T . Consider an element of length δs in the cross-section of the shaft as shown in Fig. 3.28a. The
For
the
given
values,
, , -
Fig. 3.29 Hollow circular shaft
.
37
This of course implies that if the wall thickness is constant, the shear stress is constant too. And also that that as the wall thickness thickness increases the shear shear stress decreases, and vice-versa. The nomenclature shear stress originates perhaps from the fact that there is a definite analogy between shear flow in the thin wall of a shaft and the flow of an incompressible fluid in a channel.
38
One interesting fact about this derivation is that this relation has been obtained without any reference to the material properties, and therefore, is applicable both to elastic and plastic deformations.
50
The shear stress at r = 20 mm, then, is 13.5 MPa, and at r = 16 mm is
deformed up to point A in Fig. 3.32, it follows a straight line AB with slope G, the shear modulus.
10.8 MPa. The shear stress stress varies linearly between these two values across the thickness of the shaft.
As a shaft is twisted, and the assumptions about its behaviour that plane sections remain plane and do not distort hold, the shear strain will vary linearly with the radius as shown in Sec. 3.2. The strains are then converted to stresses using the stress- Fig. 3.31 Elastic-plastic behaviour strain relation. relation. Fig. 3.32a-d show the shear stresses for various levels of straining. When the maximum shear strain is less than γY , the shear stress distribution is linear as shown in Fig. Fig. 3.32a. The maximum shear stress is less than η Y Y. As the twisting increases, the value of the maximum strain increases, the stresses increase till the stress and strain reach the
, where A is the area enclosed by the mean line of the section (here a circle of radius 18 mm). Thus, A = π (0.018 (0.018 m)2 = 1.02×10−3 m2, and = 49 N/m. The average shear stress can be obtained obtained by dividing the shear flow q by the wall thickness t to get The maximum stress is underestimated by a mere 9%.
The shear flow q by approximate procedure is given by Eq. 3.11 as
Example 3.10 tubular shaft
A
thin
rectangular
Consider a tubular shaft of rectangular section as shown in Fig. 3.30.It is subjected to a twisting moment of 100 N.m. What are the stresses in the various sides of the rectangle?
⁄
Fig. 3.30 A rectangular tubular shaft
Solution:
, where A is the area enclosed by the mean line of the section. The value of A for the given section is (22 cm – 2 cm)×(12 cm – 2.5 cm) or 0.19 m2. The shear shear flow is 0.19 m2) = 263.2 N/m. The shear flow is constant constant throughout the tube. The shear stress in the shorter (thinner) sides is q/t, or (263.2 N/m)/(0.02 m) = 13.2 kPa, and the shear stress in the longer (thicker) sides is (263.2 N/m)/(0.025 m) = 10.5 kPa. The shear flow q by approximate procedure is given by Eq. 3.11 as
(a)
(b)
(c)
(d)
Fig. 3.32 Shear stress distribution in an elastic-plastic shaft
yield values as shown in Fig. 3.32b. The slope of the stress-radius line increases as shown. Thereafter, the the plastic behaviour kicks in. The yield value of shear shear stress is obtained at a radius r = r o < R. The shear stress for r between 0 and r o increases linearly from 0 to η Y Y, and thereafter remains constant at η Y Y for r > r o. This is shown by the stress distribution of Fig. 3.32c.
3.8 Plastic deformation in torsion
40
Let us next look at what happens when a shaft is loaded beyond its plastic limit. This discussion will serve as a simple model of how problems in elastic-plastic deformations can be handled. handled. Consider a shaft made made up of an elastic-plastic material whose stress-strain behaviour can be modelled as shown in Fig. 3.31. The material is linearly elastic till the yield shear stress level39 η Y Y is reached, after which the material deforms perfectly plastically with the stress remaining constant at the yield stress value. value. If we unload such a material after after it has been
For Fig. 3.32c, the shear stress distribution is given as :
,
for
(a)
As the twisting moment increases further, the value of r o, the radius at which the yield value is reached decreases and more of the shaft is under plastic
39
Fig. 3.30 is similar to Fig. 2.29c, except that linear stress and strain have been replaced with shear stress and strain. strain. It will be shown later in Sec. 6.10 that the yield shear stress η Y Y is related quite simply to the yield stress ζ Y Y: η Y Y = ½ ζ Y Y.
40
Since the value of dθ/dz of dθ/dz is is not known, γ z is not determined, and, therefore, r o is undetermined thus far.
51
deformation. Fig. 3.32d shows the condition when the entire shaft has undergone plastic deformation. The shear stress distribution is related to twisting moment by using the same rddr at radius procedure we followed in Sec. 3.2: We take a small area element rddr at r and determine the shear force on it as ηrddr . The contribution to the twisting twisting moment of this elemental force is obtained by taking its moment about origin (by multiplying the force with radius r ) and integrating over the entire area:
∫ ∫
In the preceding section we have seen the difficulties that arise when we undertake the analysis of both elastic and plastic plastic deformations. The analysis was possible at all because we were taking a very simple case of pure torsion on a uniform circular bar.
(b)
We can however work with a simpler model in much of engineering work with complicated structures. In this this model model which which is termed as limit analysis , we take a fullyplastic stress distribution and omit the earlier elastic-plastic considerations. The load load for which the structure first becomes fully-plastic is termed as the limit load .
Example 3.11 Twist in a shaft with plastic deformation A shaft of 4 cm dia is subjected to a twisting twisting moment of 2 kN.m. Determine the twist per meter length of the shaft. The yield shear stress for steel is 125 MPa.
Solution:
Limit analysis is based on the idea that any structure made up of materials with fairly Fig. 3.33 Fully-plastic stress well-defined yield point will not undergo very distribution large deformations till all of it is in the plastic region. And, therefore, deformations will be small as long as the load is less than the limit load for for that structure. Limits load, in general, general, are much easier to calculate. Let us consider a shaft shaft in the plastic limit when the whole whole crosssection is in the plastic plastic state. The stress distribution is then then very simple, as shown in Fig. 3.33. The calculation of the limit torque torque42 for such a distribution is quite simple:
Let us first assume that only elastic deformation deformation takes place. Eq. 3.4 gives the maximum shear stress as , well above the yield value. Following the derivation and notation of the last section, let r o represent the radius at which plastic deformation is reached41. Then,
∫ ∫ ∫ ∫ ∫ ∫∫ ∫ 0 ∫ ∫ 1 0 1
⁄ ⁄
3.9 Limit Torque
If we plug in Eq. (b) the value of η of η as as given by Eq. (a), we can obtain the value of r o. Once r o is obtained, we can determine the value of dθ/dz and complete the solution of the problem.
, , -
Using these values and the fact that the shear strain is given by −3 −3 for all r , we get 1.56×10 /11.2×10 m = 0.14 −1 m . Therefore, the twist per meter meter is 0.14 rad/m.
⁄
∫ ∫ ∫
(a)
Plugging in the values of η of η Y Y = 125 MPa, T = 2.0 kN.m, and R = 0.02 m, we get r o = 0.0112 m, or 1.12 cm. The stress at this radius radius is the yield shear stress stress 125 MPa. For the value of shear modulus of 80 GPa, this gives a shear strain of 125 −3 MPa/ 80 GPa = 1.56×10 at r = 0.0112 m.
As long as the applied torque is below this value, some part of the structure will be in the elastic range and the twist will not be excessive. Limit analysis is quite extensively used, at least as a first approximation, in the design of structures.
42
The limit load for this case can be obtained quite easily from Eq. (a) in the solution of Example 3.10. The limit load corresponds to r o vanishing. Replacing r o by 0, gives the same result as here.
41
We cannot find the value of r o by using Eq. 3.2 because it has been derived under the assumption that the deformation is linear elastic everywhere.
52
Since Castigliano theorem requires that the total energy be differentiated with respect to torque applied at the point where the twist angle is to be determined, we replace the torque of 100 N.m by T . The TMD of the composite shaft, then, is as shown in Fig. 3.35. T he total strain energy is given by:
3.10 Strain energy in torsion We had introduced in Sec. 2.9 the concept of strain strain energy. The energy method based on Castigliano theorem was introduced in Sec. 2.10 where a number of examples involving axially-loaded members members were solved. The method was also applied to a statically indeterminate problem. We introduced in Sec 2.11 the calculation of strain energy for an arbitrarily loaded structure. structure. We now extend the energy methods to cases involving torsion.
6 7 and
Let us consider a shaft of radius R and of length L subjected to a pure torsional moment T. The strain energy is obtained from Eq. 2.16 using the stress distribution given by Eq. 3.4.
∫ ∫ ∫ ./ ∫ ∫ ∫ ∫
0 1 Using 100 N.m as the value of T , we get θ = 3.9 rad, or 223o, a fairly large twist. Let us, as a check, calculate the maximum shear stress in the shaft. The maximum shear stress in a circular shaft is given by . This has R, and determining the been obtained by substituting the value of I zz in terms of R η at r = R. This gives the values 47.7 MPa for the part A of the shaft, 46.6 stress η at MPa for the part B, and 63.3 MPa for the part C . These appear to be safe even as the twist appears to be excessive.
(3.12)
2 where A is the cross-sectional area and the integral of r over A has been replaced by the polar moment of area I zz.
Example spring
Castigliano theorem can be quite easily extended for the case of a torque load T to give the twist θ as:
,
3.13
Coil
Fig. 3.36a shows a closely-wound coil spring consisting of n turns of a wire of radius of r wound into a coil of radius R. Determine the elongation of the spring under the action of an axial force P, and, thereby, obtain the spring constant.
(3.13)
We now illustrate the application of Castigliano theorem.
Example 3.12 Stepped shaft Consider the stepped steel shaft shaft shown in Fig. 3.34. It consists of three parts: part A (dia 20 mm, length 1 m), part B (dia 16 mm, length 1.5 m), and part C (dia 10 mm, length 2.0 m). It is subjected to three twisting torques of 300 N.m, 200 N.m and 100 N.m as shown. Determine the total twist of the free end of the Fig. 3.34 Stepped shaft shaft.
(a)
(b) Fig. 3.36 Coiled spring
Solution: This example provides a dramatic illustration of the power of energy method and of Castigliano
Solution:
53
The twists of the two shafts are obtained by partially differentiating U with respect to T 1 and T 2, respectively. Equating the two twists, twists, we get
theorem. Fig. 3.36b shows the FBD of the spring with the wire cut at an arbitrary location. We can, by equilibrium of moments conclude conclude that the twisting moment T on the wire is clearly equal to PR, the product of the load P and the radius R of , independent of the location of the cut. Thus, the twisting moment the coil , everywhere in the wire is PR. Therefore, the strain energy due to the twisting moment in the wire constituting the spring is obtained by writing the strain energy dU of an elemental length and integrating it over the entire length of the wire, i.e., for n coils of radius R. The total energy of the spring is given by
∫ ∫
4 Here, the value of I zz,1 zz,1, the polar moment of inertia of the steel rod is πD /32 = 4 −6 4 (0.050 m) /32 = 0.61×10 m , and that of I zz,2 π (0.050 zz,2, the polar moment of inertia of 4 4 4 4 4 −6 4 Di ) /32 = π (0.080 (0.080 – 0.60 the luminium tube is π( Do – D – 0.60 ) m /32 = 2.75×10 m . The values of G1 and G2 are about 80 GPa and 26 GPa, respectively. Using the L as 0.5 m, the equivalence of twists gives: value of L
, since the total length of
−6 −6 3.90×10 T 1 = 6.99×10 T 2, or T 1 = 1.79T 2
the wire is 2πnR. 2πnR.
(b)
We can evaluate the shear force acting on the wire as V = P. These shear forces also contribute to the strain energy, but it can be shown that their contribution is of order (r/R)2 of the contribution of twisting moments, and hence can be neglected.
Solving Eqs. a and b simultaneously, we obtain the values of T 1 and T 2 as 3.2 kN.m and 1.8 kN.m, respectively. The (equal) twists in the two shafts shafts are then obtained as 0.0125 rad or 7.17o.
Use of Castigliano theorem then gives the deflection of the spring in the direction of application of force P as:
Summary
The spring constant
Example 3.14 Statically-indeterminate torsion problem Consider a composite shaft consisting of a solid steel shaft and an aluminium tube as shown in Fig. 3.37. If a torque of 5 kN.m is applied to the rigid disc attached to the left end, determine the twist of the composite shaft.
Solution: Let the torque carried by the steel rod and the aluminium tube be T 1 and T 2, respectively. Fig. 3.37b shows the FBD of the rigid end disc. From the equilibrium of this disc: T 1 + T 2 = 5 kN.m
When a circular shaft is subjected to a pure torsional load, there is no warping or distortion of cross-sectional planes, and consequently there is only one component of strain, namely γ z (and its complementary strain γz ). We first assume a twist angle θ and determine the value of the o strain γ z as rdθ/dz using rdθ/dz using purely geometrical considerations. We next convert the shear strains to shear stresses using the shear o modulus G: η z = Gγ z = Grdθ/dz. We next take a small element within the cross-section of the shaft, o write an expression for shear force acting on it in terms of the shear stresses, take its moment about the axis of the shaft and integrate over the entire cross-sectional area to obtain the twisting moment T as GI zzdθ/dz. Here I zz is the second moment of area (polar) of cross2 section, defined as integral of r dA over the cross-section. The value of I zz for a solid cylinder was found to be πD4 /32, and for a hollow shaft as . The twist θ is given by TL/GI zz. o The quantity GI zz is seen as the torsional stiffness which is defined as the torque required to produce a unit twist per meter length of a shaft. Since an area nearer the surface has a larger values of stresses that contributes more to the twisting moment than an equal area nearer the axis, hollow shafts are much more effective in resisting torsional loads. There are many situations in which the twisting moments in shafts cannot be determined by considerations of equilibrium equilibrium alone. In such shafts the geometrical constraints on twists need to be invoked to solve
(a)
Since there is no other equilibrium equation, this problem is statically indeterminate. All we have besides Eq. a is the geometric geometric compatibility condition that the twists θ1 and θ2 of the two shafts are are equal. Let us determine θ1 and θ2 by energy method. The total strain energy is given by:
54
the problems. Such problems are termed as statically indeterminate indeterminate ones. We follow the same three-step process that was invoked in the last chapter: Use equilibrium analysis to write equations for the twisting o moments in the various parts of the shaft. There may not be enough equations to determine the moments explicitly. Convert twisting moments to twists in each part. o Write geometrical compatibility conditions to complete the o analysis. For composite shafts made up of a core of one material overlaid with a sleeve of another material, the torsional stiffness is given by . The concept of shear flow q defined as the shear force per unit length of the wall was introduced for the approximate analysis of thin tubular shafts. It was shown that , where A is the area enclosed by the mean line of the wall of the shaft. Limit load refers to the load under the assumption that the whole of the structure has yielded and the shear stress everywhere is η Y If the Y. applied torque is well below the limit torque so obtained, we can presume that the twist of the shaft is not excessive.
The strain energy in torsion of a shaft is given by
.
55
For the beam to be in equilibrium, the net force or moment on the beam or on ny prt of it must be zero. Let us ‘cut’ the cntilever bem shown in Fig. 4.1a 4.1a near the wall and draw the free-body diagrams (FBDs) of the two parts as shown in Figs. 4.1b and c. Clearly, we need a force V and a moment M to balance the applied load P on the part shown in Fig. 4.1c. These are the forces and moments of internal reaction and these arise because of the deformations that are caused by the applied load. There are, of course, equal and opposite reactions on the stump of the beam built in the wall as shown in Fig. 4.1b. We can, in general, resolve the internal resistance forces and moments in to their components. Fig. 4.2 shows that that resolution. The following are the nomenclature
4 Forces and moments in beams
4.1 Introduction We have so far considered two types of structural elements: axially loaded truss elements that resisted axial forces, tension or compression, and shaft-like elements that resisted twisting moments. Beams that resist transverse loads by bending are other common common structural elements. We define a beam as a slender element (i.e., an element whose cross-sectional dimensions are much smaller
(a)
(b)
Fig.4.2 Forces and moments in a beam
and the actions of the various components:
F x:
the axial force that results in elongation (or compression, if it is negative) of the member.
F y and F z:
shear forces that result in shearing at the section. The shear shear forces are conventionally assigned the symbol V.
M z:
the axial moment that is the torsion moment that causes twisting of the member. This was the subject matter of the last chapter.
M x and M y:
transverse moments that are termed as the bending moments and cause cause the member to bend. Moment M x results in bending the beam in the y-z plane while the moment M z results in bending the beam in the x-y plane.
(c)
Fig. 4.1 A cantilever beam
than its length dimension) that essentially resists loads acting transverse to its length. Roof beams that support the roofs of buildings are beams. The horizontal deck of a road flyover flyover supported on columns columns is a beam structure. The leaf springs of an automobile suspension transfer the weight of the car to the axle (and further to wheels) through beam action. A wing of an airplane is a beam. So is the cantilever beam shown in Fig. 4.1.
We had in Chapter 2 laid out the general strategy for solving the problems of the mechanics of deformable bodies as consisting of three major steps:
56
Consideration of static equilibrium and determination of loads in various members, Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations), and Considerations of the conditions of geometric compatibility. In that chapter we had also studied the structures bearing axial loads. In Chapter 3 we considered slender members bearing twisting moments. moments. These were termed as shafts. We learnt to determine the variations of torsion moments ( M z in the notation introduced above) as torsion moment diagrams (TMDs), and then to calculate the resultant shear stresses and twisting angles of the shafts In view of the importance of the role of beam members in engineering and other structures, we devote this chapter to beams, and that too, to only the first step of the 3-step process outlined outlined above. The loads which are of immediate immediate relevance to a beam are the shear forces F y and F z and the bending moments M x and M y which are caused by the transverse transverse loads as shown in Fig. Fig. 4.3. It can be verified easily that the shear force component F z and the bending moment component M y are caused only by the vertical load P z, whereas the shear force component F y and the bending moment component M z are caused only by the transverse load P y.. This leads us to a very simple stratagem: we calculate the bending moments and shear forces for the vertical and horizontal loads separately (assuming that the other loads are absent), and then patch up the results. We shall, in the discussion that Fig. 4.3 Loads on a beam follows, consider only the vertical load system, which can easily be replicated for the horizontal loads.
Fig. 4.4 Positive shear stress and bending moment Table 4.1 Sign convention43
4.2 Sign convention It is imperative from the point of view of consistency that we define the sign convention for shear force and and bending moment at a section. Fig. 4.4 shows a beam which has been sectioned and the two resulting parts separated. The shear forces and the bending moments at the two sections are equal and opposite, as shown. We adopt the convention given in Table 4.1to ensure that shear force (or the bending moment) at either of the two sections have the same sign:
Sign of the outward normal to the section
Actual direction of the force or moment
Assigned sign to the shear force or bending moment
Along the + ve coordinate direction
Along the +ve coordinate direction
+ ve
Along the − ve coordinate direction
Along the − ve coordinate direction
+ ve
Along the + ve coordinate direction
Along the − ve coordinate direction
Along the − ve coordinate direction
Along the +ve coordinate direction
− ve
− ve
It can be verified that the shear forces and bending moments shown in Fig. 4.4 are all positive according to this convention. This convention 44 can be
43
It is interesting to note that this sign convention results in a beam bending concave upwards with a positive bending moment and convex upwards for a negative bending moment.
57
summarized graphically as Fig. 4.5. Either of these two can be used as an icon for the sign convention employed.
4.3 Loads and supports One common classification of beams is Fig. 4.5 Graphically representation of sign convention based on the kind of supports on which a beam rest. There are three idealized support systems for for beams as shown in Fig. 4.6. Support A is a termed as a pinned support . Here the beam is attached to the support in such a manner that it is not restrained from rotating about this point. point. Consequently, the reaction from the support is just a force which can be resolved in to two force components R A,x and R A,y. The support here cannot apply a moment on the beam. The support shown at point B is termed as a roller support . Here the reaction from the support can only be a vertical force R B,y. The roller does does not restrain restrain the beam from moving horizontally, and hence there is no horizontal force applied by the support. As in the case of pinned support, there is no reaction moment at a roller support as well. Beams with only pinned or roller supports are termed as simplysupported .
Fig. 4.6 Three type of beam supports and their conventional representations
We, in a similar fashion, idealize the loads that the beams carry. The left top diagram in Fig. 4.7 show a simply-supported beam with a concentrated or a point load , while the on the right shows one with a distributed load . A point load is specified as a force P with appropriate force units like N.
The support shown at C is called a built-in support. Here the beam is built into the wall so that it is restrained from rotating about point C. The beam will have a zero slope at such a point. This is a consequence of the restraining moment moment M y that the support applies to the beam at that point. point. Considerations of equilibrium will require that the reaction RC,y be equal to P, and the moment M y applied on the beam by the support will be PL, where L is the length of the beam. beam. A beam can have a built-in support at either one end or both ends. The beam shown at right in Fig. 4.6 with only one end built-in (and no support at the other end) is termed as a cantilever beam.
Table 4.2 Idealized beam supports
The lower two diagrams in Fig. 4.6 show the conventional representations of the three types of idealized supports. Please note that that these are idealizations. Most practical supports are more complicated, but many can be analysed quite satisfactorily by replacing them with one one or the other of these idealizations. The dashed lines in these represent the (exaggerated) deflected shapes of the beams. Note that the cantilevered beam has zero slope at the built-in end. Table 4.2 summarizes these idealized supports.
44 Please note that many current books use a sign convention entirely different from this. It is, therefore, advisable that the sign convention convention used be noted in each problem that you solve. The sign convention can be denoted denoted easily by a graphical symbol similar to the ones shown in Fig. 4.5
58
Support type
Freedom of motion
Reactions present
Built-in support
No degree of freedom
A moment as well as horizontal and vertical reaction forces
Pinned support
Single degree of freedom - rotation.
Horizontal and vertical reaction forces
Roller support
Two degrees of freedom – freedom – Only vertical reaction rotation and horizontal force movement
A distributed load, on the other hand, is specified by a load density w as force per unit length with appropriate units like N/m. What is shown on the beam on right
A beam could also be subject to an externally-applied bending moment as shown in Fig. 4.8b. Here too, the dashed line represents the (exaggerated) deflected shapes of the beam.
4.4 Determining shear forces and bending moments Determination of shear forces and bending moments at any point in a beam is rather a straight forward procedure. The procedure consists of taking a section of the beam at that point to expose the shear force V and the bending moment M at that point and drawing a free body diagram (FBD) of one portion of the beam. We will first solve a couple of simple examples to illustrate the procedure and then draw up a generalized scheme of doing things in more complicated situations.
Example 4.1 A cantilever beam with a load at the free end
Fig. 4.7 Two idealized loadings of beam
Consider a cantilever cantilever beam as shown in Fig. 4.9. It is loaded at the free end. Draw the shear force diagram (SFD) and the bending moment diagram (BMD).
is idealized as a uniformly distributed load, wherein the load density w per unit length does not change along the length of the beam. However, in some situations, the load density could vary, as is shown in Fig. 4.8a. Here the load is distributed, but the loading density varies along the length of the beam. It is modelled as a linearly varying distributed load . If the load density is w N/m at the end of the beam of length L, the load density at any x from the left end is given simply by wx/L N/m.
(a)
Solution: We first calculate the reactions at the support. A simple equilibrium analysis shows that the end reaction is a force P upwards and a clockwise moment PL as shown. We next calculate the shear force V and the bending moment M at an arbitrary location x from the support. For this purpose we take a section at that location and draw the FBD of the
(b) Fig. 4.8
(a) A simply-supported beam with non-uniformly distributed loading (b) A simply-supported beam with a concentrated applied bending moment
59
Fig. 4.9 SFD and BMD for a cantilever beam with end load (a) A cantilever beam with support reactions (b) FBD of the section beyond x beyond x (c) Shear force diagram (SFD) (d) Bending moment diagram (BMD)
right-hand portion45 of the beam as Fig. Fig. 4.9b. The shear force and the bending moment on this section are the external loads acting on this part of the beam and will be shown on the FBD . We do not know their directions. directions. In such a case we show them assuming they are positive. positive. Since the exposed face on which they are negative x-direction, the positive shear force acting has an outward normal in the negative x must be downward and the positive bending moment must be clockwise (as shown).
beam there and drawing drawing the FBD. It should be obvious that there are two kinds x is less than L/ 2 or greater than L/ 2. of FBDs depending on whether the value of x 2. Fig. 4.10b shows the FBD for x < L /2. We shall show the SF and BM as external loads on this FBD. We do not know their directions. directions. As before, we show them assuming they are positive. Since the exposed face on which they are acting has has an outward normal in the positive x-direction, the positive shear force must be upward and the positive bending moment must be counter-clockwise (as shown).
46
A simple equilibrium analysis of this FBD gives, for x < L /2:
A simple equilibrium analysis of this FBD gives:
∑ ∑ :
or
:
∑ ∑
(a)
, or
:
(b)
or
:
Since the location of the section x is arbitrary, and changing the value of x does not change either the FBD or Eqs. (a) and (b), these equations give the shear x between 0 and L. The resulting force and the bending moment at all values of x values have been plotted as shear force and bending moment diagrams in Figs. 4.9c and 4.9d, respectively. These show the variations of the respective respective quantities with x.
, or
(a) (b)
Thus, for x < L /2, SF has a constant value of – – P/ 2 while the value of the BM increases linearly from 0 at x = 0 to a value of PL/ 4 at x = L /2.
Example 4.2 A simply-supported beam with a concentrated load in the middle Consider a simply-supported beam with a concentrated load at the mid-point as shown in Fig. 4.10a. Draw SFD and BMD for this beam.
Solution: The first step in the solution of this problem is determining the reactions at the support. Drawing an FBD of the entire beam (or using symmetry) symmetry) we can easily find the two vertical reactions to be P/ 2 each. To determine the shear force force (SF) and the bending moment (BM) at any location x (measured from the left support) of the beam, we need to externalize the SF and BM by taking a section of the 45
We could have chosen either pat of the beam but have chosen to draw the FBD for the right-hand portion of the beam because this results in an FBD with fewer unknown forces simplifying the equations. 46 M, x x in Eq. (b) denotes that moments have been taken about a point at x. This is convenient because the moment contribution of the unknown force V about this point is zero. It may further be noted that in the equilibrium equation, the sign to be used with the bending moment is the standard sign convention for moments: positive for clockwise and negative for counter-clockwise. counter-clockwise. Even though the bending moment moment shown is positive (negative on a face with negative outward normal), it is taken as positive in the equilibrium equation.
Fig. 4.10 SFD and BMD for a simply-supported beam with a concentrated load in the middle (a) (b) (c) (d) (e)
60
Loading diagram FBD when the section is taken in left half of the b eam FBD when the section is taken in right half of the beam Shear force diagram Bending moment diagram
x > L /2, the generic FBD changes This formulation is valid only for x < L /2. For x and is as shown in Fig. 4.10c. Here the applied load P too figures in the FBD.
Introduce the unknown shear force V and the bending moment M at the cutting plane. These should be shown assuming they are positive positive
An equilibrium analysis of this FBD gives, for x > L /2:
∑ ∑ :
or
:
,
(c) or (d)
Thus, for x > L /2, SF has a constant value of + P/ 2 while the value of the BM decreases linearly from PL/ 4 at x = L /2 to a value of of zero at x = 0. The resulting 47 SFD and BMD are plotted in Figs. 4.10d and 4.10e, respectively.
4.5 General procedure for drawing shear force and bending moment diagrams by method of sections On the basis of the method adopted in the last two examples, the following generalized procedure may be formulated:
Draw an idealized loading diagram of the beam. Determine the reactions at all supports. If the reactions cannot be determined, the beam is statically indeterminate and further progress cannot be made without considering the the deflections of the beam. These will be introduced in Chapter 7. Determine the number of segments with distinct loading pattern to cover the entire beam. In Example 4.2, we needed two different different types of FBD: one without the concentrated load P, and the other with it. In practice, this means that we segment the beam such that the end of a segment is at the location of a discontinuity discontinuity in loading pattern. These could be the concentrated loads or moments, or where the type of distributed loads changes.
Fig. 4.11 SFD and BMD for a simply-supported beam loaded symmetrically with two concentrated loads (a) (b) (c) (d) (e) (f)
For each of the segments identified above, introduce a cutting plane (at a location x from the left end) and draw an FBD of either part of the beam (as convenient). In example 4.1 we had drawn the FBD of the right portion (since it carried fewer loads), while in Example 4.2, we drew the FBD of the left portions (since these permitted simpler expressions for force moments).
Loading diagram FBD when the section is taken in the left third of the beam FBD when the section is taken in the middle third of th e beam FBD when the section is taken in the right third of the beam Shear force diagram Bending moment diagram
according to the sign convention introduced in Table 4.1 above. Thus, in Example 4.1 where we had drawn the FBD of the right-hand part of the beam, the cut face has the outward normal in the negative x-direction and, therefore, the positive SF is downwards and the positive BM is clockwise. This is exactly opposite of the case of Example 4.2 where the outward normal is in the positive x-direction and, therefore, the positive SF is upwards and the positive BM is counter-clockwise.
47
Since the BM is positive, the beam bends concave upwards, as is physically apparent from the loading
61
∑ ∑ or
Determine the expressions for SF and BM by equilibrium considerations, equating the sum of vertical forces and the moment (about a convenient point) to zero.
, a constant
(d)
And from the equilibrium analysis of the FBD of Fig. 4.11c, we get for x > 2 L /3: :
Plot the resulting expression to obtain the shear-force and bending moment diagrams (SFD and BMD). These diagrams are conventionally conventionally drawn exactly beneath the loading diagram of the beam as in Figs. 4.9 and 4.10.
or
:
(e)
,
We give below a few more examples to illustrate the general procedure.
Example 4.3 A simply-supported beam with two concentrated loads Consider a simply-supported beam of length L with two loads acting at L /3 and 2 L /3 from the end as shown in Fig. 4.11. Draw the SFD and BMD of this beam. beam.
Solution: As usual, we begin with with determining the reactions reactions at the supports. Through a simple equilibrium analysis (or by using the symmetry of the problem) the reaction at either support is a force P upwards.. It is easy to see that we need to segment the beam in this case in three different ways. One kind of FBD results when the section section is taken for with no loads acting on it (as in Fig. 4.11b). Another kind results when the section is taken for a value of x one external load P x between L /3 and 2 L /3. In such an FBD one acts at x = L /3 (as in Fig. 4.11b). And the third kind results when the section is taken for a value of x between 2 L /3 and L. In such an FBD two external loads act, one at x = L /3 and the other at x = 2 L /3 (as in Fig. 4.11c).
We next show the unknown shear force V and the bending moment M at the newly exposed sections assuming they are positive and using the sign convention of Table 4.1. In each of the three FBDs, the positive SF is upwards and the positive BM is counter-clockwise, since all outward normals are are along the positive x-axis.
Fig. 4.12 SFD and BMD for a simply-supported beam with uniformly distributed load
From the equilibrium analysis of the FBD of Fig. 4.11b, we get for x < L /3:
∑ ∑ :
or
:
(a) (b) (c) (d) (e)
(a)
, or
(b)
Similarly, from the equilibrium analysis of the FBD of Fig. 4.11b, we get for L/ 3 >x > 2 L /3:
∑ ∑ :
or
:
Loading diagram Typical FBD for a segment of the beam Distributed load replaced by a statically equivalent l oad Shear force diagram Bending moment diagram
or
(c) ,
62
(f)
The resulting SFD and BMD48 have been plotted as Figs 4.11e and f.
Solution: The total load on this beam is wL. The symmetry of the beam suggests49 that the reaction at each support is wL/ 2. 2. Since there is no no discontinuity in the loading with no concentrated load or moment, only one kind of FBD as drawn in Fig. 4.12b suffices for this problem. The presence of distributed load poses a problem in determining the moment moment contribution of the loading. loading. To this, we replace the distributed load with its statically equivalent load 50 as shown in Fig. 4.12c. It consists of the total load wx acting at the centroid of the distributed load, which is at the mid-point of the beam, i.e., at a distance of x /2 from the left end.
Example 4.4 A simply-supported beam with uniformly distributed loading Consider a simply-supported beam with a uniformly distributed load of w N/m as shown in Fig. 4.12. Draw the SFD and BMD for this beam.
From the equilibrium analysis of the FBD of Fig. 4.12c, we get for all values of x:
∑ ∑ :
or
(a)
Thus, the shear force varies linearly, with a value of at x = 0, and a value of at x = L. It is zero at x = L /2. The variations of shear force force are plotted as SFD in Fig. 4.12d. :
or
,
(b)
This represents a parabola with M = 0 at x = 0 and L, and a maximum value of at the mid-point x = L /2. These variations of bending moment along the length of the beam are plotted as BMD in Fig. 4. 12e.
Example 4.5 A simply-supported beam loading with a concentrated moment at the middle Consider a simply-supported beam of length L loaded in the middle with a concentrated moment M 0 as shown in Fig. 4.13a.Draw the shear force and bending moment diagrams for this beam.
Fig. 4.13 SFD and BMD for a simply-supported beam with a concentrated moment (a) (b) (c) (d) (e)
Loading diagram Typical FBD for a section in first half of the beam Typical FBD for a section in second half of the beam Shear force diagram Bending moment diagram
49
This can also be obtained by drawing the FBD of the whole beam and replacing the distributed load by the statically equivalent load system consisting of the total load wL acting at the mid-point of the beam. A statically equivalent load is the alternate load which produces the same net force and moment at any point. 50 It can easily be verified that the loading shown in Figs. 4.12b and 4.12c are statically equivalent as defined above.
48
Here too, the BM is positive and the beam bends concave upwards, as is physically apparent from the loading.
63
The deflected (but exaggerated) shape of the beam is shown as the broken line in Fig. 4.13a. As noted earlier, the beam deflects concave concave upwards where the BM is positive and convex upwards where it is negative.
Solution: We begin with determining determining the reactions at the supports. supports. Through a simple equilibrium analysis the reaction at the left support is determined as a force equal to M 0 /L, while t the right support it is − M 0 /L. The negative sign indicated that 51 the reaction at the right support is directed downwards, as shown in t he figure .
4.6 The area method of drawing the SFDs and BMDs
We need to segment the beam in two different ways: one, when the section is taken at (without including the concentrated moment M 0 in it) acting on it (as in Fig. 4.13b), and the other when which included the concentrated moment M 0 in the resulting FBD shown as Fig. 4.13c.
The method of sections outlined in the last section is a simple method of determining shear forces and bending moments, but it can get quite tedious in all but very simple cases. We give here an alternate method known as the area method.
We next show the unknown shear force V and the bending moment M at the newly exposed sections assuming they are positive and using the sign convention. In either FBD, the positive positive SF is upwards and the the positive BM is counter-clockwise since the outward normal in either case is along the positive xaxis. From the equilibrium analysis of the FBD of Fig. 4.13c, we get for x < L /2:
∑ ∑ :
or
:
, or
(a) (b)
Thus, the shear force V is constant in this part, while the bending moment M increases linearly from 0 to at the mid-point of the beam. Similarly, from the equilibrium analysis of the FBD of Fig. 4.13c, we get for L/ 2 >x > L:
∑ ∑ :
or
:
or
(c)
,
(d)
The shear force V is constant in this part too, with the same value as in the first part. The bending moment M shows an interesting behaviour: it jumps down from to at x = L /2 nd then increses, gin linerly, from − at the mid-point to 0 at the e nd-point of the beam52. The resulting SFD and BMD are plotted as Figs. 4.13d and e.
51
It has been assumed here, for simplification, that at least one support is not restraining the beam horizontally, so that there is no horizontal reaction force. Anyway, the presence of a horizontal force will not affect the SFD and BMD. 52 Attention is drawn to and interesting feature of SFDs and BMDs as is apparent to the five examples given so far: the line depicting shear force has a discontinuity wherever there is a concentrated force as a load, and the BMD is
Fig. 4.14 Elemental FBDs for three different kin ds of loads
now seen to be discontinuous at the location of the concentrated moment. We will elaborate on this later.
64
Consider the beam shown in Fig. 4.14, which carries a concentrated load P at location B and a concentrated moment M 0 at location C as shown. A part of the beam (segment DE ) is subject to a distributed load. The density q( x) of this load measured in N/m may vary along the length.
We last consider an elemental section at a point within the segment DE which carries a distributed load. The FBD of such a section is shown as Fig. 4.14e. 53
For x within the beam segment DE, the equations of equilibrium give:
∑ ∑ :
Let us determine how the SF and BM change with x along the beam. beam. Let us first take a small element of length dx at a location x within the segment AB where no load acts on the beam. We draw an FBD of this element of the beam as shown in Fig. 4.14b. Here we have shown shear force V and bending moment M at location x, and shear force V + dV and bending moment M + dM at location x + dx. It can be verified that all directions have been shown such that SF and BM at either location are positive. The application of equilibrium to this FBD gives: gives:
∑ ∑ :
, or
:
:
(b)
∫ ∫
This implies that the SF as well as BM does not change along a beam segments which does not carry any load. We next consider an elemental section of the beam which includes point B, the location of the concentrated load. Fig. 4.14d shows the FBD of this element. element. The application of equilibrium to this FBD gives: , or
:
(c) , or for (d)
infinitesimal dx,
, or
:
(h)
(4.1) (4.2)
Let us first consider Eqs. (a), (c), (e), (g) and 4.1, all pertaining to changes in shear force V . Eq. (a) states that that V does not change across any element which does not carry any load. Eq. (c) suggests that at the location of a concentrated load, the shear force jumps up by a value equal to the load (acting downwards). Eq. (e) states that V does not change across any element carrying a concentrated
Fig. 4.14d shows the FBD of an element of the beam which includes point C, the location of the concentrated moment M 0. The application of equilibrium to to this FBD gives: :
, or
The eight equations (a) – (h) above (along with Eqs. 4.1 and 4.2) provide a framework for an very convenient method for drawing the FBDs and BMDs of the beams directly without drawing FBDs of different segments..
These indicate that at the location of a concentrated load, the shear force jumps up by a value equal to the load (acting downwards), but there is no change in the value of the bending moment. It can be deduced quite easily that had the load P been acting in the upwards direction, the SF would have jumped down by the same value.
∑ ∑
(g)
The area method gets its nomenclature from these two equations which say state that the change in SF across a beam segment is equal to the area under the loading curve of that segment, and the change in BM across a beam segment is equal to the area under the shear force curve of t hat segment54.
∑ ∑ :
, or
This suggests that for a distributed load, the rate of change of shear force at a location is equal to the (downward) load density at that point. And, similarly, the rate of change of bending moment is equal to the (negative of) shear force at that point. We can recast the equations (g) and (h) as:
(a)
, or
or
(e)
53
Two things should be noted here. First, the contribution of the distributed load load to the moment equation has been obtained by replacing the distributed load by its statically equivalent concentrated load qdx at the midpoint of the element, i.e., at location dx /2. Second, this contribution is negligible negligible since this moment moment is of , while the other terms are of first order only. second infinitesimal order , 54 The difference in signs in Eqs. 4.1 and 4.2 is due to the fact that the load density q has been defined as positive downwards. If we had taken the usual sign convention that upward forces are positive, we would have got a negative sign before integral in both the equations.
, or for infinitesimal dx,
(f)
Thus, at the location of a concentrated moment, the bending moment jumps down by a value equal to the applied moment, but there is no change in the value of the shear force across such an element.
65
moment, and Eq. (g) says that the rate of change in shear force is equal to the density of loading.
the (positive) applied applied moment. Eq. (h) states that that the rate of change in bending moment is equal to the (negative of) shear force at that point.
In fact, Eq. 4.1 summarizes all this information quite succinctly if we assume a concentrated load as the integral of the load density at that point. The change in shear force across a segment of the beam is equal to area under the loading density curve.
Eq. 4.2 contains all of the above points. points. The change in bending moment while traversing a segment of the beam is equal to the area under the shear force curve (but with a reversed sign) over that segment. The following procedure may be adopted for drawing the BMDs:
From these we can deduce the following procedure to draw the SFD:
2.1. First complete the shear force diagram of the beam.
1.1. First draw the loading diagram of the beam. 1.2. Calculate the reaction at the supports.
2.2. Start at a point slightly left of the left support where the bending moment is taken as zero.
1.3. Start drawing the SFD at a point slightly left of the left support where the shear force is taken as zero.
2.3. Travel along the beam to the right, modifying the BM using the rules given below.
1.4. Travel along the beam to the right, modifying the SF using the rules given below.
2.4. The slope of the BMD at any location is equal to (negative of) the SF value at that location 2.5. Change in BM between any two locations is equal to the (negative of) the area under the SF curve. 2.6. Concentrated moments cause a jump in BMD at the location of the moment. 2.7. Go up for every negative concentrated moment, down for every positive concentrated moment.
1.5. Shear force does not change over the segment of the beam with no external load. 1.6. The presence of a concentrated moment does not change the value of shear force. 1.7. A concentrated force causes a jump in SFD at the location of the force: go up for every load downwards, down for every load upwards.
2.8. Proceed till you are slightly right of the right support where the BM should again be zero again. This last is a check on calculations.
1.8. The slope of the SFD at any location is equal to the distributed load density (load per unit length). Positive slopes for loads downward, and negative slopes for loads acting upwards.
This is known as the area method of drawing the BMDs. It can be verified quite easily that the BMDs Figs. 4.9d, 4.10e, 4.11f, 4.12e, and 4.13e follow from the respective shear force diagrams using the above rules.
1.9. Change in SF between any two locations is equal to the area under the distributed load curve. Positive changes for negative areas and negative changes for positive areas.
The following example illustrates the simplicity of the method of area.
Example 4.6 Beam with an overhang
1.10. Proceed till you are slightly right of the right support where the SF should be zero again. This last is a check on calculations.
Consider a 4 m long simply-supported beam with an overhang as shown in Fig. 4.15. The overhang portion AB carries a distributed load of density 20 kN.m. There are two concentrated loads at locations C and E as shown. It also carries a concentrated counter-clockwise moment of magnitude 20 kN.m at location D. Draw the SFD and BMD for this beam.
This is known as the area method of drawing the shear force diagram. It can be verified quite easily that the SFDs Figs. 4.9c, 4.10d, 4.11e, 4.12d, and 4.13d follow from the respective loading diagrams using the above rules. We, in a similar fashion, can deduce the rules for drawing the BMDs from Eqs. (b), (d), (f), (h) and 4.2, all pertaining to changes in bending moment M . Eq. (b) states that M does not change across any element which does not carry any load or moment. Eq. (d) implies that at the location of a concentrated load too, there is no change in the value of the bending moment. Eq. (f) states that at the location of a concentrated moment the bending moment jumps down by a value equal to
Solution: We first determine the reactions R1 and R2 at the supports. With reference to the loading diagram of Fig. 4.15, we can write the equilibrium equations. Equating the vertical forces to zero, we get:
66
or
. /
We start at zero at the left end. We first encounter the uniformly distributed distributed load. Using rules 1.8 and 1.9 above, the shear force would increase at a constant the rate of 20 kN/m, and after one meter it would have increased from 0 at A to 20 kN at B, as shown. A concentrated load (acting upwards) upwards) occurs at the location location B. Therefore, following rule rule 1.7, the value of shear force jumps jumps down by 70 kN R1) from 20 kN to − 50 kN. Since there is no lod on segment BC, (the value of R the value of SF through this segment remains constant constant (rule 1.5). At location C, the sher force jumps up from − 50 kN to − 10 kN (rule 1.7), and thereafter remains constant at this value across segment CE, since the this segment does not have any load except a concentrated moment at D, and rule 1.6 says that SF does not change due to the presence of a concentrated moment. moment. The downward load at E makes the SF jump up through 40 kN from – 10 kN to + 30 kN (rule 1.7 again). The shear force remains constant constant at this value through segment segment EF , and then jumps down to 0 at F because of the presence of the concentrated load R2 = 30 kN. The final value of 0 completes the check (rule 1.10).
, (a)
We next take the moment of all the forces about the left support and equate it to zero to get55:
,
(b)
From Eq. (b) we get
Using this value in Eq. (a) we get
We are now ready to draw the shear force diagram.
It should be obvious that this method of area is so much simpler than the method of sections where we would have needed to draw five different kinds of FBDs and, then, write ten equilibrium conditions for those five FBDs. After completing the SFD, we can draw the the BMD too. We start at zero for a point slightly left of point A. The SF increases linearly from 0 at A to 20 kN at B. By rule 2.4, then, the slope of the BD chnges linerly from zero to − 20 kN·m/m. The linear change in slope implies implies that the curve is of second order in x, that is, it is parabolic, as shown. The change in the value of the bending moment in going from A to B is the area under the SFD in this segment (rule 2.5), which is . This area is positive and, therefore, the change in BM is – is – 10 10 kN.m, giving a value of BM at point B as – as – 10 kN.m. The shear force across segment BC is constant at – at – 50 50 kN, so the slope of the BM curve there is constant at + 50 kN.m/m, and the value of the BM through this section of length 1 m changes (linearly) through 50 kN.m, the area under the SF curve in this segment (rule 2.5). The value of BM at point C is 40 kN.m.
– 10 kN, and therefore the BM curve slopes up at From point C to D, the SF is – 10 this rate and the value of BM changes from 40 kN.m kN.m to 45 kN.m in 0.5 m. The presence of a positive external concentrated moment of 20 kN.m at point D makes the BM jump down to 25 kN.m at this point (rule 2.6), after which the curve continues its upward march at a rate of + 10 kN.m/m to acquire a value of 30 kN.m at point E . The SF in the last segment EF is constant at +30 kN. Therefore, the BM there decreases linearly from a value of 30 kN.m to 0 in a length of 1 m. The end value of 0 indicates that the calculations are in order.
Fig. 4.15 Drawing SFD and BMD usin g area method
55
Here again we take the moment of the distributed load by replacing it with the statically equivalent load, which is the total force (20 kN)·(1 m) acting at the centroid (of this load) which is located at 0.5 m from the left end of the beam.
67
Example 4.7 A built-in frame Consider the frame ABCD shown in Fig. 4.16a which is built in at A. Draw the SFDs and BMDs of the various segments of the frame.
Solution: To begin the analysis, we need to consider the FBDs of the various sections of the frame. Fig. 4.16b shows the upper segment DC of the frame. There will be a force and moment acting at the point C due to the restraining action of the segment CB. Using the requirements of equilibrium we evaluate the vertical force as P upwards and the moment as PL /2 clockwise. Using the area method outlined above, we can easily obtain the SFD and BMD of this segment as shown in Fig. 4.16b. The FBD of the segment CB is shown in Fig. 4.16c. The load at point point C of this segment is equal and opposite to that at point C of the segment DC . Using equilibrium of this segment we obtain the force at point B as P upwards, and the moment there to be PL /2 clockwise. The force in this segment segment is axial and, and, therefore, there is no shear force anywhere. anywhere. The BMD has been obtained using the rules 2.1, 2.6 and 2.8. The positive sign of the bending moment moment indicates that this segment will bend concave upwards (viewing from right), which agrees with what is expected intuitively. We then move on to the cantilevered section AB, which carries a force P downwards and a counter-clockwise moment PL /2 at point B as shown in Fig. 4.16c. We can obtain the the reaction force and moment moment at end A by equilibrium considerations. Due to the upward force P, the SFD jumps down to – to – P at the left end A and remains so up till the other end where it jumps down to 0. The BMD jumps down to − PL /2 at point A due to the presence of a concentrated moment PL /2, and thereafter increases at a constant rate of P because of the presence of the shear force – force – P throughout this length reaching a value of + PL /2 at the end B. It jumps back to zero there because of the presence of a counter-clockwise moment PL /2.
Example 4.8: An optimization problem Consider a simply-supported beam of length L carrying a load 2P at the middle as shown in Fig. 4.17a. It also carries a load P at each of the two ends. The two supports are located a distance a inwards from the ends. Determine the value value of the distance a required to minimize the maximum bending moment in the beam.
Fig. 4.16 Drawing FBDs and BMDs of th e various segments of a frame
68
value of bending moment at the mid- point is [−Pa + P( L/ 2 − a)], or +P( L/ 2 − 56 2a). We can continue in the same fashion and complete the BMD. The magnitude of the maximum negative bending moment is Pa, and the magnitude of the maximum positive moment is P( L/ 2 − 2a). As the value of a increases, the first value increases while the second second decreases. A little reflection should convince the reader that the minimum of the maximum value will occur when these two magnitudes are equal:
,
or
, or
.
Thus, we minimize the maximum value of the bending moment by locating the supports at L /6 from either end of the beam.
Summary Study of the behaviour of beams is very important because the beams are central to the design of most structures. We, in this chapter have have concentrated on determining the variations in shear forces and bending moments along the length of the beam. Fig. 4.17
We first introduced the types of supports and the loads that the beams carry. Three idealized supports were identified:
Solution:
From the symmetry of the beam (or from the two equilibrium equations for the whole beam) we determine the reaction force at either support to be 2P.
We first draw the SFD using the area method. We start the SFD slightly left of the left end of the beam with a zero value. A downward force P is encountered at the left end which causes the SFD to jump up to a value of P (Fig. 4.17b). Since there is no other loading till the left support, the SF remains constant at P till then. The reaction at the left support is 2 P upwards which causes the SFD to jump down by 2P to a value of – – P at the left support. We continue in a similar manner and complete the SFD as shown.
The built-in support which allows no degree of freedom to the beam and admits both components of the reaction force as well as a reaction moment. The pin support which allows one degree of freedom, the freedom to rotate about that point. This support admits admits both components components of the reaction force but does not admit a reaction moment A roller support which restraints the beam only from the vertical motion and, therefore, admits only the vertical reaction force component.
We next introduced the sign convention for the shear force and bending moment. The sign is based on two parameters: the direction of the outward normal at the section, and the direction of the force or the moment itself. If both of these are in the positive coordinate directions, or both in the negative coordinate directions, the sign of SF or BM is taken taken as positive. And if one of them is positive positive and the other negative, the sign is taken as negative. negative. This is summarized in Table 4.1 and as icons of Fig. 4.5.
The BMD is also drawn drawn in the same fashion (Fig. 4.17c). We start at zero. In the first segment of length a, the SF is constant at a value of +P, and therefore, the BM in this section decreases at the rate of P per unit length. length. The change in value till the left support is the areas under the SFD of this segment, which is Pa. Thus, the vlue of B t the left support is −Pa. In the second segment segment (of length L /2 – /2 – a) the vlue of SF is − P and, therefore, the slope of bending moment curve here is +P. The change in value of BM is P( L/ 2 − a) from −Pa. Thus, the
The method of sections for drawing the SFDs and BMDs uses the following procedure:
56
69
Or use the symmetry of the beam.
The following process for drawing BMD was concluded:
Draw an idealized loading diagram of the beam.
Determine the reactions at all supports.
Determine the number of segments with distinct loading patterns to cover the entire beam.
For each of the segments identified above, introduce a cutting plane (at a location x from the left end) and draw an FBD of either part of the beam (as convenient).
Introduce the unknown shear force V and the bending moment M at the cutting plane.
Determine the expressions for SF and BM by equilibrium considerations, equating the sum of vertical forces and the moment (about a convenient point) to zero.
The other method of determining SFDs and BMDs is the method of areas is based on the two equations:
, and . From the integration of these we concluded that the change in SF across a beam segment is equal to the area under the loading curve of that segment, and the change in BM across a beam segment is equal to the area under the shear force curve of that segment. The following process for drawing SFD was concluded:
First draw the loading diagram of the beam.
Calculate the reaction at the supports.
Start drawing the SFD at a point slightly left of the left support where the shear force is taken as zero. Shear force does not change over the segment of the beam with no external load. A concentrated force causes a jump in SFD at the location of the force: go up for every load downwards, down for every load upwards. The slope of the SFD at any location is equal to the distributed load density (per unit length). Positive slope for loads downwards and negative slope for loads acting upwards. Change in SF between any two locations is equal to the area under the distributed load curve. Positive changes for negative areas and negative changes for positive areas.
70
First complete the shear force diagram of the beam. Start at a point slightly left of the left support where the bending moment is taken as zero. The slope of the BMD at any location is equal to (negative of) the SF value at that location Change in BM between any two locations is equal to the (negative of) the area under the SF curve. Concentrated moments cause a jump in BMD at the location of the moment. Go up for every negative concentrated moment, down for every positive concentrated moment.
consider a very simple case of a beam subject to pure bending in Fig. 5.1.
57
as shown
To obtain the relations between the bending moment M b and the stresses and strains we shall follow the strategy that we outlined in Chapter 3 while dealing with the the case of torsion of a shaft. This was the reverse of the generalized strategy outlined outlined in Chapter 2. The reason for this again is the same as in the case of torsion: unlike the situations involving axial loading treated in Chapter 2, it is not possible to make the assumption of uniform stresses in the case of bending of beams. Before going further, let us determine the nature of deformations in beams under pure bending and the nature of stresses that result from such deformations. Fig. 5.2 shows a beam with a grid drawn on it. This beam is subjected to a positive bending moment, and bends as shown. It is clear from the enlarged portion shown in Fig. 5.2c that the beam elements
5 Stresses in beams
5.1 Introduction In problems involving bending of beams we may be interested in determining:
The stresses that result when a beam carries the specified loads. We may consequently determine the maximum load that a shaft can carry without failure. The deflection of a beam when carrying the specified loads.
We have seen in the last chapter that almost all loads in beam result in variation of the resulting bending moment and shear stress along its length. This inevitably introduces complications in the determination of the behaviour of the beam. To obtain a Fig. 5.1 Beam in pure bending foothold on the problem we first
57
Fig. 5.2 Bending stresses in beams (a) Undeformed beam (b) Beam after bending (c) Enlarged portion showing that there is contraction near the top and extension near the bot tom
The bending moment can be constant along the length of a beam only if the shear force V is zero everywhere. This is possible only i f there is no other load on the beam except concentrated moments at the two ends.
71
near the top surface of the beam are under compression, while those near the bottom are in elongation. As a consequence, there are compressive compressive stresses near the top and tensile tensile stresses near the bottom. bottom. It should be clear that since there in no net tension (or compression) in the beam, the forces due these stresses must total out to zero. The bending moment at the section is the integrated moment of the elemental forces due to these stresses.
assumptions are valid in this case of pure bending of a beam of uniform and symmetrical (about a vertical axis) section:
The strategy for this problem then consists of:
First using symmetry considerations to establish the nature of deformation in the beam under pure bending. Using the geometric considerations to determining the tensile strain across the beam section as a function of the vertical coordinate. This is the macro to micro conversion discussed in Section 2.1. We next use the material properties to convert these strains to stresses. This is a micro to micro transformation. In the last stage we convert the stresses into bending moment, a micro to macro conversion . We use the fact that the the total bending moment is the given M b, and the total tensile force at the section is zero. This completes the solution to the problem.
Calculation of strain from the geometry of deformation
We now consider the geometry of deformation further to obtain an expression for strain. It was argued above that the beam elements near the top are under compression while those near the bottom are under tension. It stands to logic, then, that there must be a plane in the beam (near the middle) which is neither under tension nor compression. Let AB represent such a plane for the beam segment 1-2-3-4 of Fig. 5.4. This plane with its its extension to the rest of the beam is termed as the neutral plane of the beam. It should be understood that we do not, as yet, know where this is located, except for the fact it must be somewhere between the top and the bottom surfaces of the beam.
5.2 Relating curvature of the beam to the bending moment Consider a beam with a crosssection that is laterally symmetric (i.e., it has symmetry about a vertical axis) subjected to a pure bending moment M b as shown in Fig. 5.3. Under the action action of this this positive bending moment, the beam bends and acquires a curvature. The coordinate axes used here are shown in the figure. By using symmetry arguments we can show that the following
The beam bends in the shape of a circular arc. This follows from the argument that a uniform bending moment acting along the length of the beam must result in a uniform curvature. Plane cross-sections of the beam remain plane after the bending. Thus, plane sections 1-2 and 3-4 of the undeformed beam of Fig. 5.4 remain plane in the deformed shape as well. Initially parallel sections (sections 1-2 and 3-4 of the undeformed beam of Fig. 5.4) must deform so that they have a common point of intersection as shown. This common point of of intersection is the centre of curvature of the deformed beam.
Let us consider a plane CD a distance y up from the neutral plane AB. The undeformed lengths of
Fig. 5.3 A symmetrical beam
72
Fig. 5.4 Geometry of deformation of a b eam
planes (along the x-axis) AB and CD are equal to begin with. Since the length of the plane AB is unchanged (this being the neutral plane), the undeformed length of the plane AB or CD is seen as from Fig. 5.4b, 58 where ρ is the radius of curvature of the neutral plane of the beam. The deformed length of the plane CD is . Thus, the contraction contraction in in the length of this plane located a distance y up from the neutral plane is given by ydθ. The strain at this location is, thus, given
like that shown shown in Fig. 5.5. The trace zz of the neutral plane is termed as the neutral axis. Note here we have shown the neutral axis to be nearer the bottom plane of the beam. beam. This results in the top plane plane of the beam to be farther away from the neutral axis (NA) than the bottom plane is. Consequently, the maximum compressive stress is larger than the maximum tensile stress in the beam.
Since there are no shear strain components, there will be no shear stresses. Further, there are no other longitudinal longitudinal stresses, ζ yy or ζ zz, in this 60 case of pure bending .
(5.1)
Converting loading
It can be seen that in this simple case of pure bending (i.e., with no loading other than the bending moment which is constant along the length of the beam) there are no shear shear strain components. However, there will be longitudinal strains ε yy and ε zz due to the presence of ζ xx through Poisson ratio.
Conversion of strain strain to stress is simple. simple. The tensile stress component can be obtained by multiplying the strain with the elastic modulus E : (5.2)
Note that we know neither the value of y (because the location of the neutral plane has not yet been established) nor the value of ρ, the radius 59 of curvature . But we have two conditions that we shall shortly use. The distribution of the stresses across the section is
into
Consider an elemental area dA in the right face of the beam as shown in Fig. 5.6. The force acting on this elemental area due to the bending stress is ζdA acting along the normal to the Fig. 5.6 Calculating loading at a section section as shown. Note that that we have shown this force outwards from the area, assuming ζ to be tensile (i.e., positive). positive). The algebra below will automatically automatically take care of the sign if the stress is compressive. The neutral plane of the beam is 61 shown as grey . The z-axis is the neutral axis of the section under consideration.
Converting strain to stress
stress
The total axial force on the beam can be found by integration as
∫
. Substituting the value of ζ using Eq. 5.2, we get
Fig. 5.5 Stress distribution on a section of the beam
where y is the vertical distance measured from the neutral axis. But there is no axial force F x in this case of pure pure bending. Therefore,
58 The radius of curvature ρ curvature ρ is ∞ for a flat beam. As the beam bends, the value of ρ decreases. We introduce the term curvature for 1/ ρ. ρ. Curvature is usually κ (read kappa). The value of κ denoted by κ (read of κ increases increases as the beam bends. It is 0 for a flat beam (corresponding to ρ to ρ = ∞). 59 But we have two conditions that we have not used so far. The total axial force on the section is zero, and the total moment acting on the section is M b.
73
60
We can now determine the lateral shear strains ε yy and ε zz as
61
Note again that we do not as yet know the location of the neutral plane.
.
∫
The bending formulae (Eqs. 5.4 and 5.5) are sometimes summarized
(5.3)
What does this imply? It simply defines the location of the neutral axis from which the distance y is measured. Thus, the neutral axis of a beam is located such that the first moment of area about that axis is zero .
63
as:
(5.6)
It may be noted that the equation above have been obtained for the case of pure bending, i.e., for the case of constant bending moment in the absence of any other load. The results for the more general case are very difficult difficult to obtain. But wherever these have been obtained, it is seen that for slender members, the resulting stress distribution is quite close to those obtained above. It is for this reason that we use the results obtained here in the general case as well.
Recall that the centroid of an area is defined as the point about which the first moment of area is zero. zero. Therefore, the neutral axis of a beam passes through the centroid of the cross-section 62. We next calculate the moment (about the z-axis) of these elastic forces and equate it to M b, the bending moment on the beam:
Since the values of are given by
∫
The integral is recognized as the second moment of area I zz about the z-axis, which, in this case, is the centroidal axis. This is a geometric quantity. Once the shape of a cross-section is known, the value of I zz can be calculated. calculated. See Appendix A to learn learn some important important properties of the second moment of area and the procedure for its calculation.
and
are zero, the lateral strains
and
(5.7)
Thus, the normal strains in the plane of the cross-section are proportional to the axial strains, but of opposite opposite sign. Further, since ε xx varies with y, the vertical distance measured from the neutral axis, the strains ε yy and ε zz 64 change across the cross-section. cross-section. This leads to the the deformation of the section as shown in Fig. 5.7. Notice from Eq. 5.6 that the curvature of the beam 1/ ρ is given by
We can, therefore, write an expression for 1/ ρ, the reciprocal of the radius of curvature (which is termed as the curvature κ = 1/ ρ) of the beam as
(5.4) The product is termed as the bending rigidity of the beam. The more
the value of the bending rigidity, the less is the curvature κ of the beam (for a given value of the bending moment), and more is the radius of curvature ρ.
63
Notice the similarity of this with a similar equation that we can obtain for the case of torsion of a shaft from Eqs. 3. 2 and 3.3:
.
We can combine Eqs. 5.2 and 5.4 to obtain
Note that in this equation for shafts, the z-axis is the longitudinal axis, unlike in the present case where the z-axis is a transverse axis while the longitudinal axis is represented as the x-axis.
(5.5)
These stresses are termed as the bending stresses.
64 Note, in particular, that the neutral axis (which is the trace of the neutral surface in the section) section) is now curved. This transverse curvature of the crosssection of the beam is termed as anticlastic curvature. The neutral plane now has double curvature, one in the x-y plane, and the other in the y-z plane.
62
We had chosen the location of the neutral plane in Fig. 5.6 closer to the bottom of the beam anticipating this result.
74
see straight away that there will be an upward reaction of 200 kN at the support. From the moment balance we obtain obtain the moment at support support to be 300 N.m, counter-clockwise as s hown. We draw the SFD by the method of areas. We start at 0, jump down 200 N at the left support (load being upwards), continue at ‒ 200 200 N till the first load is encountered. The downwards load makes the SF jump jump up to ‒ to ‒ 100 100 N at 1 m. m. The SF jumps up again at the second load.
Fig. 5.7 Deformation of the shape of a rectangular beam (a) The undeformed rectangular section (b) variation of longitudinal stress distance y from the neutral axis (c) variation of longitudinal strain ζ xx with distance y across y (d) variation of lateral strains ε yy and ε zz with y with y (e) the deformed ε xx across y shape of the cross-section (deformation (deformation exaggerated). Note that the neutral axis which is the trace of the n eutral surface in the section) is curved.
The BMD is drawn in the same fashion. We start at 0, jump down to ‒ to ‒ 300 300 N.m at the support because of the presence of the positive reaction moment. Through the first one meter of the beam, the BM line line has a positive slope of 200 N.m/m (equal to the negative of the constant shear force). The value of BM at 1 m, the midpoint midpoint of the beam, is ‒ 100 100 N.m. The change in value of BM between x = 0 m to x = 1 m is equal to the area under the SFD over this length. From x = 1 m to x = 2 m, the slope of the BD is +100 N.m/m, the SF being constnt t ‒100 N over this length.
The quantity EI zz can be interpreted as the stiffness of the beam. The more its value, the less is the curvature of the beam, i.e., less is the bending.
We, thus, see that the maximum bending moment occurs at the root of the cntilever where its vlue is ‒300 N.m. N.m. This, then, would be the loction loction of the maximum maximum stresses. Since the BM is negative, the beam beam would bend convex up. The curvature 1/ ρ given by Eq. 5.4 will be negative.
We give below a few examples of calculation of bending stresses.
Example 5.1 Maximum stresses in a cantilever beam Consider a 2 m long cantilever MS beam of section 10 mm × 20 mm loaded as shown in Fig. 5.8. Determine the location and magnitude of the maximum bending stress.
The value of I I yy needed in this equation has been obtained in Appendix B 3 (Eq. B.6) as bh /12, where b is the width of the section (10 mm in this I yy is case), and h is the height of the section (30 (30 mm). Thus, the value of I ‒ 3 ‒ 3 ‒ 8 3 4 (10×10 m) × (30×10 m) /12, which is 2.25×10 m .
Solution: We need to obtain the bending moment distribution first. For this, we follow the procedure introduced in the last chapter, and first calculate the reactions at support. From the balance of vertical forces we can
The maximum tensile stress occurs at the top of the beam ( y = 0.010 m) at the root ( x = 0), and is given by Eq. 5.5 as
= 200 MPa
The value of the maximum compressive stress will be the same, except that it will be at the bottom of the root of the beam. The beam is loaded almost to its ultimate strength. Note that given a bending moment, the stresses are independent of the material of the beam. A beam with lesser elastic modulus will bend more more (giving larger values of curvature 1/ ρ and strains) but will produce same stresses.
Fi . 5.8 A cantilev cantilever er beam beam
75
It is interesting to compare these results with the case when the beam section has the longer side side horizontal. Notice that the length length of the horizontal side now is a factor of 3 larger, the vertical side a factor of 3 smaller, the value on I zz as now two factors in 3 smaller. The maximum bending stress that varies like y/I zz is 3 times larger, and the radius of curvature that varies like I zz is two factors in 3 smaller, i.e., the curvature th (1/ the beam section section is vertical. vertical. These results are ρ) is 1/9 of when the summarized in Table 5.1.
Example 5.2 Bending stresses in an angle-section beam Consider a simply supported angle-section steel beam loaded as shown in Fig. 5.9. Its section is the same as was used in Example B.5. Determine the maximum tensile and compressive stresses in the beam.
Solution: To draw the SFD and BMD, we calculate the reactions at the supports. We determine the reaction at the right support by writing the balance of moments about the left support. This reaction is found to be 50 N. Then, by vertical force balance, the reaction at left support is found to be 250 N as shown.
Table 5.1 Comparison of two different ori entations of beam sections Section orientation
Factor
b
10 mm
h
30 mm
I zz
‒ 8
2.25×10
30 mm 10 mm 4
m
‒ 8
The SFD of the beam can be drawn drawn easily by the method method of areas. The SFD jumps up through 100 N at the left-end because of the presence of a downward force, then jumps down through 250 N to ‒150 N due to the concentrated reaction at 1 m, jumps up through 200 N to +50 N because of the downward load at 2 m length, and then jumps down to zero because of the 50 N upward reaction at the right support.
3 1/3 4
0.25×10 m
1/32
ζ xx
200 MPa
600 MPa
3
Radius of curvature ρ (at the point of maximum bending moment)
15 m
1.67 m
1/32
The BMD of the beam starts at zero and has a slope of 100 N.m/m through the first 1 m length. length. The value at 1 m is equal to the (negative of) area under the SFD over this length, length, which is 100 N.m. We complete the BMD in the same fashion following the rules given in Sec. 4.6.
It is clear that when more material of the beam is away from the centroidal axes (on either side of it), more is the value of the second moment of area, and the stiffer is the beam.
There are two extrema in the BMD: a negative 100 N.m value at x = 1 m, and a positive 50 N.m value at x = 1 m. The centroid of the cross-section of the beam can be determined by taking the first moment of the area about an axis passing through the bottom of the section as was done in Example A.5 in Appendix A where the vertical value of it was found to be 16.1 mm. The second moment of the area about the centroidal axis was also determined there. The strategy was to divide the area into two rectangles, rectangles, and then for each rectangle first determining the value of I about its own centroidal axis, and then using the parallel-axes theorem to transfer it to the neutral axis, i.e., the centroidal centroidal axis of the full section. The value was ‒ 7 4 determined as I zz = 2.99×10 m . We can next determine the stresses. At location x = 1 m where the mximum negtive bending moment of ‒100 N.m acts, the beam bends convex up so that the maximum tensile stress acts at the top plane which
Fig. 5.9 A simply supported angle-section beam
76
has a value of y = +33.9 mm measured from from the neutral axis. Similarly, the maximum compressive stress act at the bottom plane which has a y = +16.1 mm measured from the neutral axis. value of y
Example 5.3 Comparison of rectangular, T-section and I-section beams Consider three crosssections of beams shown in Fig. Fig. 5.10. 5.10. The three sections have equal areas and, hence equal weights per unit length of the beams. Compare the maximum tensile and compressive stresses in the three beams for a bending moment M b.
We use Eq. 5.5 to determine the stresses. These are arranged in the Table 5.2 given below. At location x = 1 m where the bending moment is ‒100 N.m, the maximum tensile stress (at y =0.0339 m) is given by:
The maximum compressive stress occurs at the bottom plane at y = ‒0.0161 m:
We can similarly determine the stresses at the location x = 2 m where the bending moment is +50 N.m. These are shown in the table above.
Bending moment, M moment, M b
1m
2m
‒ 100 100 N.m
+50 N.m
̅ ∑ ̅ ∑
Next, we calculate the the second moment of inertia. inertia. For this, again, we divide the cross-section in a number of rectangles, calculating the value of I for each rectangle about its own centroidal axis, and the transferring it to the neutral axis using the parallel-axes theorem. Thus,
∑ . ̅ /.
At top plane ( y y = 0.0339 m) 11.3 MPa
Maximum compressive stress
At bottom plane ( y y = ‒0.0161 m) ‒ 5.4 5.4 MPa
(c)
The calculations have been organized organized in Table 5.3. We first calculate the location of the neutral axes by calculating the y-coordinates of the centroids of the three sections. sections. This is determined by dividing dividing the areas into the requisite number of rectangles and then using the formula: .
Stress distribution
Maximum tensile stress
(b)
Solution:
Table 5.2 Stresses at the locations of maximum positive and negative bending moments x-location
(a)
Fig. 5.10 Three beam sections of equal areas
At bottom plane ( y y = ‒0.0161 m) 2.7 MPa At top plane ( y y = 0.0339 m) ‒ 5.6 5.6 MPa
77
The stresses are then evaluated from Eq. 5.5. These stresses vary linearly from zero at the neutral neutral axis. The stresses are compressive above above the neutral axis and tensile below below it. In each case, the maximum compressive stress occurs in the top plane, while the maximum tensile stress occurs at
5.3 Composite beams Consider the case of pure bending of a composite beam made up by bonding two materials of different elastic properties, in particular, of different elastic modul moduli. i. Let us consider a section with symmetry symmetry about yaxis as shown in Fig. 5.11. Let this beam be subjected to pure bending bending with a positive bending moment M b. Let us determine the moment curvature relations for this beam following the procedure used in Sec. 5.2.
Table 5.3 Comparative stresses in beams of equal weight
Section Location of NA from the base I zz
30 mm
38 mm
30 mm
5.40×10 ‒ 7 m4
5.21×10 ‒ 7 m4
6.77×10 ‒ 7 m4
Using the symmetry of the situation, we can again argue that plain sections will remain plain and that the beam will bend in a circular arc. The geometry of deformation will be the same as depicted in Fig. 5.4, a nd we would get an expression for the linear strain similar to Eq. 5.1
Stress distribution
(5.8)
where ρ is the curvature of the neutral plane, i.e., a plane that does not have any strain, and y is the distance measured from the neutral plane. y for max tensile stress
‒ 30 30 mm
‒ 38 38 mm
‒ 30 30 mm
y for max comp. stress
+30 mm
+22 mm
+30 mm
max tensile stress Max comp. stress
4
5.56×10 M b Pa 4
5.56×10 M b Pa
4
4.43×10 M b Pa
4
4.43×104 M b Pa
7.29×10 M b Pa 4.22×10 M b Pa
Note that, as before, for a given curvature of the beam, the strain varies linearly the distance from the neutral plane and does not depend on the material of the beam.
4
However, when we continue to the next step of converting to stress, the material properties come in and the stress is given by
the bottom most plane. It should be noted that in the case of a T-beam, the maximum compressive stress is much smaller than the maximum tensile stress since 65 the neutral axis is closer to the compressive side .
(5.9)
where E i stands for the elastic modulus of the specific material at that location. Note, in particular, particular, that while the strain at the junction of two materials is continuous, the stress shows a discontinuity (see Fig. 5.12). Note that the strain, as well as the stress, is zero at the neutral plane. The strain increases linearly on either side of the neutral axis, compressive 66 above it and tensile below it . The stress distribution distribution is discontinuous. discontinuous. The stress jumps at the location location where the material changes,. changes,. The strain, being a purely geometrical quantity in a beam, shows no such jump. Consider two points close together on either side of the interface of the
Another thing to note is the reduction in the maximum stress in the I I zz, which in turn results from section. This is a result of a larger value of of I more of the section area (compared to the rectangular section) being away 2 from the central axis. Since an are’s are’s contribution in I zz is weighted by y , the areas far away from the centroidal axis contribute more, making the beam stiffer.
65 But the total compressive force is exactly equal to the total tensile force on the section, there being no net axial force. There is more area of the section on the compressive side than on the tensile side. It is for this reason that the neutral axis has shifted upwards causing a reduction of tensile stresses.
66
The bending moment, being positive, introduces positive curvature (concave upwards) resulting in the strains as shown.
78
two materials. Since the strains are the the same, the changed value of E E will result in different values of stress on the two sides. 67 Hence the jump . The location of neutral plane and the value of ρ ρ are as yet undetermined. To evaluate these we, as before, use the considerations of equilibrium. equilibrium. The total
or, or,
,
.
We now consider the integral over the whole area as a sum of a number of integrals, each over an area of uniform material. Thus, (a) (b) (c) Fig. 5.12 (a) beam beam section (b) strain strain distribution (c) stress distribution
force on the section is evaluated as by integration as Substituting Substituting the value of ζ using Eq. 5.9, we get
∫ ∫
∫
∫ ∑ ∫ ∑ ∫ ∑ ∫ ∑ .
Similarly,
represents the location of the centroid of the i system. From these
̅ ∑∑ ̅
.
th
,
where
area in we
coordinate get (5.10)
68
This gives the location of the neutral axis within the section .
,
To obtain the moment curvature relation, we write the equilibrium condition for the moments:
where E i is the elastic modulus of the material specific to the area element dA (and y is the distance of the area element measured from the neutral plane) .
This axial force should vanish, and therefore, the neutral plane should be located such that
∫ ∫ ̅ ̅ ∫ ∫
The last integral is again interpreted as the sum of integrals over areas of different (homogeneous) materials to obtain . Then, the curvature κ = 1/ ρ can be seen as
∫
∑
.
∑
To interpret this equation, consider Fig. Fig. 5.13 5.13. The distance y is measured from the neutral axis. Let distance y’ be measured from an arbitrary axis from which the neutral axis is located at a distance of . Then , and the equation of equilibrium becomes
̅
̅
(5.11)
where I zz,i is the second moment of the area Ai about the neutral axis of the section.
∫ ̅
Once the curvature is known, we can calculate the strain as
,
68
It may be noted that in Eq. 5.10 the areas of different materials are weighted by the elastic moduli of the materials. These happens because because this relation is obtained from the balance of force equation where forces are determined from the stresses which are E times the strain. The strains, by themselves, are geometric quantities. A consequence of this is that the neutral plane shifts towards the area with the larger modulus of elasticity.
67
Two things may be noted in Fig. 5.12. First, near the interface the value of stress in the upper material is less than that in the lower material. This indicates E for the upper material is less than that for the lower material. that the value of E Second, note that all the straight lines depicting the stresses in the various materials originate from zero value at the neutral axis. Why?
79
and
∑ ∑
(5.12)
moment of the lower area about its own axis + are a× (distance between the neutral axis and the centroid of 2
area 1)
(5.13)
3
= (0.08 m)×(0.02 m) /12 + (0.08 m)×(0.02 m)×(0.04 m –
where E is the elastic modulus of the relevant material.
0.01 m)
Example 5.4 Stresses in a sandwich beam
Similarly,
2 ‒ 6
Consider a beam of sandwich construction in which two steel plates are separated by a block of lightweight rigid foam material with a negligibly small elastic modulus Fig. 5.14 A sandwich beam compared to that of steel (Fig. (Fig. 5.14). Find the stress distribution for a bending moment of 100 N.m.
moment of the upper area about its own axis + area× 2 (distance between neutral axis and the centroid of area 3) 3
= (0.08 m)×(0.02 m) /12 + (0.08 m)×(0.02 m)×(0.04 m – 0.07 m)
2 ‒ 6
Then,
‒ 4
= 1.68×10
‒ 4
‒ 1
1.68×10 m )
, where A1 refers to the
area of the lower steel plate with E 1 = 200 GPa and = 10 mm, A2 refers to the area of foamy material with a negligibly small vale of E, and A3 to the area of the upper plate with E 3 = 200 GPa and . Using these values, we get
̅ ∑
‒ 1
m
The value of the maximum compressive stress is then determined from Eq. 5.9 with E = 200 GPa, and y = 0.04 m,
It is easy to see from the symmetry of the section that the neutral plane is situated at the middle. We can also use Eq. 5.10 to obtain the location of the centroid.
4
= 1.49×10 m .
Solution:
̅ ∑∑ ̅ ̅̅̅
4
= 1.49×10 m .
= ‒1.34 ,
The minus sign sign indicating indicating that it it is a compressive compressive stress. similarly determine the maximum tensile stress (at y = ‒ 0.04 m) s +1.34 MPa.
We can
The rigid foam serves the purpose of shifting the two steel areas away from the centroid increasing the Fig. 5.15 The stress distribution in sandwich beam contribution of each area to I zz, thereby increasing the stiffness of the beam. If the foam was not separating the two steel plates, the total height height of the section would have been 40 mm and the value of I zz would have ‒ 6 3 4 been (0.04 m)×(0.04 m) /12 = 0.213×10 m , much less than that of the composite beam with the resulting decrease in beam stiffness, increase in curvature and increase increase in the maximum value value of stresses. It is always
0.40 m
The curvature is given by Eq. 5.11
, E 2 being negligibly small.
Here refer to the second moment of the three areas about the neutral axis (passing through the middle of the section as shown above). 80
̅ ∑∑ ̅
beneficial to keep the section areas far away from the axis. It was for the same reason that I-beam of Example 5.3 was better than the rectangular section.
5×π×(0.020 m) 2 /4, and for concrete: E c = 20 GPa, m, and Ac = (0.450 m)×(0.200 m). Using these values in Eq. 5.10, we get
Example beam
Note that as expected, the stiffer material in the lower half of the section brings the neutral axis down (from 225 mm for the concrete alone).
5.5
Reinforced
m.
concrete Fig. 5.16 Reinforced
Reinforced concrete construction is a very concrete beam section ingenious use of two materials with complementary properties. Common concrete is a brittle material which has good strength in compression (about 20 MPa). Structural steel on the other hand has good strength a tension (about 200 MPa). All beams have have compression compression on on one side and tension on the other. This suggests that we combine the two materials to produce beams in which steel is loaded in tension while concrete is Fig. 5.17 Stress distribution loaded in compression.
∑
The stiffness of the composite beam is obtained from (refer Eq. I zz’s re obtined by first evluting the I ’s 5.11). The values of I ’s bout the centroidal axis of the concerned area and, then, transferring it to the neutral axis using the parallel-axes theorem.
64 5 7 4 5 ∑ 4
m
in a reinforced concrete beam
Fig. 5.16 shows the section of such a reinforced concrete beam. What is the maximum bending bending moment this beam can transmit?
Solution:
4
m
Let us take the value of E c, the elastic modulus of concrete as 20 GPa, while that for steel, E s as 200 GPa.
Therefore, the stiffness of the beam is
We will use the following strategy to solve this problem:
=
3.86×107 N.m2,
1. Locate the neutral axis using Eq. 5.10. 2. Determine the stiffness of the beam using Eq. 5.11. 3. Assuming a bending moment M b, calculate using Eq. 5.13 the maximum compressive stress in concrete and maximum tensile stress in steel. 4. Equate these stresses stresses to the the specified limiting stresses. The minimum of the two maximum moments will be the answer to the problem.
And the curvature is
The maximum compressive load will occur in the top plane of the beam which is (450 ‒ 205) mm wy from the neutrl plne. From Eq. 5.13,
We first determine the location of the neutral axis for the section using Eq. 5.10. Here, for steel: E s = 200 GPa, m, and As =
=
Pa, and
81
and therefore, it must be ‒ Pd Pd at point B. The BM over portion AB has, thus, a Pd . This constant value of ‒ Pd would require a clamping moment of + Pd. The resultant BMD must be as shown in Fig. 5.19. 5.19. It can be verified verified from the loading diagram that this indeed is in agreement with the other loads on the strip.
Pa
For a maximum permissible compressive stress of 20 MPa in concrete, the maximum permissible moment is given by , or M b = 157.6 kN.m. For a maximum tensile stress of 200 MPa in steel, the maximum permissible moment is given by , or M b = 249 kN.m. kN.m. The minimum of these two, then, is the permissible permissible maximum moment that this reinforced concrete beam can sustain.
Example 5.6 Force required to bend a strip
Now we are ready to evaluate the load P in terms of the given parameters. R. We know the radius of curvature of the strip over the length AB s ‒ The negative sign arises from the fact that the beam is bending convex
Consider a strip of length l, width b and height h clamped to a block with a circular arc surface of radius R as shown in Fig. 5.18. On application of a force P to the free end, a length c of the strip AB) comes in contact with the ( AB block. Find the force P.
Solution:
Fig. 5.19
upwards. Using Eq. 5.4, we get
, or .
5.4 Stresses in beams carrying shear forces We have so far considered the case of a beam in pure bending only. The only stresses that exist in such a beam are tensile stresses, termed as the bending stresses because they arise from Fig. 5.20 Relative sliding in a stack of bending of the beam. In a more planks acting as a beam general case, there are other stresses as well, notably shear stresses. To demonstrate the plausibility plausibility of the presence of shear stresses, consider the simply-supported stack of planks69 shown in Fig. 5.20. 5.20. The planks will exhibit exhibit relative sliding sliding as shown. If these planks were glued together, they would tend to slide and the glue will be in shear trying to hold the planks back from sliding. If it were a solid beam, this would cause shear stresses to be present which tend to hold the different layers together.
Fig. 5.18
We are given that the length AB of the strip has a constant radius of curvature. curvature. This implies that the bending bending moment over this length is constant, which, in turn, implies that the shear stress on this part of the strip vanishes and that there is no other loading on this part. Let us consider consider the shape of the the SFD of this strip. The SF over the length AB is constant at zero. There is no load load over the portion portion BC . This implies that SF over this segment segment too is constant. This constant SF will jump up through P at the end C because of the presence of a concentrated load P (downwards) at that that point.. But the shear force must must be zero after end C. This is possible only if the SF is constant at +P over the segment BC. Since there no SF over AB, and the SF over BC must be +P, there must be a concentrated reaction at point B where the strip loses contact with the the radius block. block. The resultant SFD must be as shown in Fig. 5.19.
69
This stack of five planks will have five times the moment carrying capacity of a single plank. But if we had used a beam of the combined combined thickness, the moment carrying capacity would would have been 25 times! Can you show this?
The BM at the free end C must be zero. The presence of a negative SF over the length BC implies that the BM is increasing over this segment, 82
Consider next the simplysupported beam loaded in the middle as shown in Fig. 5.21. The SFD and BMD of the beam have also been shown in the figure. It is clear that the beam will have shear forces acting over its entire length. The presence of the shear forces will cause the Fig. 5.21 bending moment to vary along the length as shown. In fact, this follows follows from the fact that the slope slope of the bending moment line is equal to (negative of) the shear force value.
development of shear stresses on the bottom surface of the slice acting towards right as shown. This shear is η yx with a negative sign (using the sign convention of stresses introduced in Fig. 5.23 Complementary Chapter 2) . We had seen earlier earlier (Sec. 1.7) shear stress on the left face that shear stresses occur in complementary pairs: . Therefore, a stress component component acts on the exposed x-face of the element (Fig. (Fig. 5.23). The forces due to these shear stresses acting on this face give the resultant shear force V at the section.
5.5 Relating shear stresses to the shear force in a beam The discussion above leads us to the strategy that we adopt to relate shear stresses to the shear force force in a beam. We take a slice of the beam of length dx extending from a distance y (from the neutral axis) to the top surface of the beam (see Fig. 5.24a). Let the bending moment on the left face of this element be M, while that on the right face be M + dM (Fig. 5.24c).
The presence of a positive bending moment will cause the beam to bend with a positive curvature resulting in compressive stresses in the upper part and tensile stresses in the lower lower part of the beam. beam. But since the moment varies along the length of the beam, the bending stresses (at the y) also varies with x. Thus, if same value of y we take a slice of the beam (like the shaded portion in Fig. 5.21), the bending stress distributions on its two ends are entirely different resulting in a mismatch of horizontal forces.
The bending stresses due to these bending moments at the two locations, x and x +dx (and vertical location y’ from the neutral axis) are then given by:
, and
Fig. 5.22 FBD of a slice of the beam of Fig. 5.21
This fact is illustrated by drawing a free body diagram of this shaded portion as shown in Fig. 5.22. 5.22. The (compressive) bending stresses stresses on the right end vary linearly in the vertical vertical direction. These would be given by equation
The net axial forces on the two sections are:
∫ ∫
,
, and
where y is measured from the the neutral axis of the section. There are no stresses on the left end since the bending moment moment there is zero. The bending stresses alone then give rise to a net horizontal force to the left. This force tends to slide this slice to the left which results in the
where the integration is to 83
be carried out on area A’ , which is the area of the elemental section from y upwards. Note that at x, the positive ζ is directed in the negative xdirection (the outward normal there being in the negative x-direction) and, therefore, F ,x (the force, by convention, being positive in positive coordinate direction) is obtained as the negative of the integral of stresses. However at x + dx, the positive stresses are in the positive coordinate direction and, hence, F ,x+dx is the integral of the stresses on that face.
Here V is the shear stress, Q is the first moment (about the neutral axis) of above y, b is the width of the section at location y, and I zz the section area above y, is the second moment of the whole area. It can be shown easily that we can write the following expression for Q in terms of the distance of the centroid of area A’ from the neutral axis (Fig. 5.25):
̅
∫ ̅
The presence of a shear force on this element causes the bending moments on the two sides of the element to be different, resulting in different values of of the axial forces on either either side. The net axial force is given by
(5.16)
The following examples illustrate the procedure for calculating the shear stress distribution distribution across beam se ctions.
.∫ /
Example 5.7 Shear stress distribution across a uniformly loaded cantilever beam
where y’ is the distance measured from the neutral axis and, thus, can be replaced with y. The integral above above is the first moment moment of area A’ about the neutral axis, where the area A’ is the area of the section above the location where we need to determine the shear stress (see Fig. 5.23). This is given the symbol Q.
Consider a cantilever beam loaded uniformly with a load intensity of 100 N/m as shown in Fig. 5.26. The beam has a rectangular section of width 10 mm and depth 30 mm. Find the distribution of shear stresses at the root section.
(5.14)
Solution:
This difference in axial forces is balanced by the shear stresses acting on the bottom surface of the element element (Fig. 5.24d). If we assume the average stress on this surface to be η yx (= η xy), the net shear force on this face is η yx×(b×dx ), where b is the width of the section at y. Since η yx is positive in the negative x-direction, this force is in the negative direction. Writing the axial force equilibrium of the FBD of Fig. 5.24d, we get
We first determine the reaction at the built in support. By vertical force balance the vertical reaction is determined determined to be 200 200 N upwards. To determine the reaction moment at root, we replace the distributed load by its statically-equivalent concentrated load of 200 N at the centroid of the distributed load, which is at 1 m point. The moment balance equation then gives the reaction moment as 200 N.m counterclockwise.
,
or
Vdx (realizing from Eq. 4.2 that We replace dM by by ‒ Vdx
) to obtain
To draw the FBD, we start at zero. The shear force jumps down 200 N due to
Thus, the average shear stress is given by
Fig. 5.25
(5.15) 84
Fig. 5.26 A uniformly loaded cantilever beam and the corresponding SFD and BMD
the concentrated force at x = 0. Thereafter, the SFD increases at a constant rate of 100 N/m due to the uniformly distributed load of the same density, till it reaches the value of 0 at the free end of the beam.
Fig. 5.27 shows graphically the shear stress distribution section.
Example 5.8 Shear stress distribution on wide-flanged I-beam Consider a simply-supported overhang beam with uniformly distributed load as shown in Fig. 5.29. 5.29. The beam section is a wide-flanged I71 section . Obtain the variation of shear shear stresses in the beam section.
Solution: As with all statically-determinate beams, we start with the loading diagram of the beam, calculate the reactions at the support and draw the SFD of the beam using the method of areas. Because of a uniformly distributed load of 500 N/m, the SFD has a line Fig. 5.29 A simply-supported overhang beam with uniformly sloping up at the distributed load same rate. But
a rectangular section
The maximum shear force is V = ‒200 N nd occurs t the root of the bem. bem. The sher stress t distnce
. Because the section of 3
the beam is rectangular ( b =10 mm, h = 30 mm), its I zz is given by bh /12 ‒ 8 4 or 2.25×10 m , and the NA is located at 0.015 m from the top. top. The value of b, the width of the section at a distance y from NA is constant at
∫ ̅
0.010 m. The value of Q is obtained from Eq. 5.16: . Here A’ represents the area above the location y (shaded area in Fig. 5.26). Here , and therefore,
̅ . / .
The
shear
stress,
then,
70
is
Let us compare the magnitudes of the shear stresses with those of the bending stresses in this simple case. The maximum bending stresses are of order order , while the maximum shear stresses are of order . Therefore, . The maximum value of Q in this case is (b.h/ 2). 2).h /4, and so . It is reasonable to suppose that M is of order V.L, where L is the length of the beam, and therefore, . This ratio is small for slender beams. 71 The horizontal portions away from the central axis are termed as flanges while the central vertical portion of a beam is termed as web. A wide-flanged I-beam is also termed as a W-beam
.
It is a parabolic distribution with the maximum at y = 0, i.e., at the neutral axis, and going to zero at the top and the bottom surfaces.
85 Fig. 5.28 Distortion of a rectangular
across the beam
This shear distribution results in a similar shear strain distribution across height of the beam. Fig. 5.28 shows an exaggerated view of the deformation of the beam under the action of shear forces.
To calculate the shear stress, we do not need to calculate the bending moments. However, we do so here just to to illustrate again the the area method of drawing the BMDs. After starting at 0, the BMD jumps down to 200 N.m because of the positive reaction moment of the same magnitude. Then, due to a negative shear force which is constantly decreasing in magnitude, the bending moment increases with a linearly decreasing rate, i.e., the BMD has a parabolic shape as shown in Fig. 5.26. Total change in BM (as per Eq. 4.2) over the length of the beam is equal to the area under the SF curve over the same length. This is 200 200 N.m, and therefore, the BM at the free end Fig. 5.27 Shear stress distribution on turns out to be 0, as it should be.
y from the NA is given by Eq. 5.16 as
70
the presence of the two reactions (750 N each) makes the SF line jump down by 750 N at the two supports. supports. The resulting SFD is as shown.
at y = 0, and is given by
. The value at y = 0.01 m is obtained ‒ 6 3 by using 6×10 m for the value of Q, and is obtained as 0.77 MPa.
The BMD has also been drawn in Fig. 5.29 even though we do not require it for the purposes of this example. The maximum positive shear force of 375 N magnitude occurs just before the two supports, while the maximum negative magnitude occurs just after the supports. The maximum shear stresses will also occur at these same locations. locations. We calculate calculate the shear stresses stresses using Eq. 5.15. We require for this purpose the value of I zz. This is obtained obtained by dividing dividing the section area into three rectangles A, B, and C , as shown in Fig. Fig. 5.30. The symmetry of the section ensures that the centroid of the section is in the middle.
- , , ‒ 7
Fig. 5.33 shows the distribution of shear stress η xy across the section. It may be noted that the web of the section is the primary shear-bearing element in a wide- flanged beam. The value of the stress η xy in the flanges is quite low.
4
m.
To calculate the value of Q across the section, Fig. 5.30 I-section divided up first consider the shaded area at a distance y in three rectangles from the NA as shown in Fig. 5.31a. The value of Q for this is given by or valid for y between 0.010 m to 0.020 m. Thus, the value of Q across the flange varies ‒ 6 3 parabolically from 6×10 m at y = 0.01 m to 0 at y = 0.2 m.
, , -
5.6 Shear flow in beams It was shown above that the shear stress η xy in the flanges of the wide-flanged Ibeam of the previous example is quite low. But it can be shown that the flanges carry η xz as well. To evaluate η xz, consider an area at one corner (at a distance z from the vertical axis of symmetry) as shown in Fig. 5.34. The FBD of Fig. 5.34 An element element and its FBD to determine the this area is also shown. shear stress η xz Take a small area element dA in the back face of the section. Let this area be located at a distance y from the the NA. The tensile tensile stress acting on this area where the bending moment
For y between 0 and 0.010 m, the shaded area (Fig. 5.31b) is divided in two parts. The contribution of the top part is clearly the value of Q for y = y)(0.010 0.01 m, and the contribution for the second part is (0.01 m – ‒ 6 2 3 y)/2], or (0.5×10 ‒ 0.005 y ) m . Thus, the value of Q m)×[ y + (0.01 m – in the web is , which is again parabolic, with -6 3 ‒ 6 3 a value of 8.5×10 m at y = 0 and 6×10 m at y = 0.01 m. Fig. 5.32b shows the variation of Q.
The shear stress η xy can now be determined from Eq. 5.15. At the location of the maximum shear force (= 375 N), the maximum shear occurs (a)
(b)
(c)
Fig. 5.32 (a) The I-section, (b) the variation of Q, and (c) the
(b) Fig. 5.31
In calculating this last value, we have used 0.01 as the value of b, the width of the section. A value of 0.77 MPa is actually the value of shear stress at a point slightly below y = 0.01 m where we can take the value of b to be 10 mm. At a point slightly slightly above above this point, the width of the section suddenly increases by a Fig. 5.33 Shear stress η xy factor of 4 to 0.040 m and, therefore, there is a distribution in an I-beam discontinuity discontinuity in the value of shear stress which decreases by a factor of 4 to 0.19 MPa. This has been shown in Fig. 5.32c.
We use the parallel axes theorem to calculate I zz.
or I zz =2.93×10
(a)
86
72
is M cn be tken to be ‒ My/I zz, where I zz is the second moment of the section My/I zz).dA. The total force on the back re. The totl force on this element is ‒( My/I face is then the integral of this over over the area of the face. We can , in a similar fashion, calculate the force acting acting on the front face of the area. It will be exactly the same, except that M will be replaced by M + dM, and that this force is backwards. The net force due to bending stresses on the two faces is the difference of the two, i.e., the integral over the area of (dM/I zz) ydA. The integral ydA is the first moment of area which has been given the symbol Q. Thus, the of ydA net force due to bending stresses is QdM/I zz. The change in moment, as before, can be replaced by ‒ by ‒ Vdx, Vdx, where V is the shear force at the section.
tips and is maximum at the junctions with the web . It is pertinent to point out here that the shear flow in bending has been obtained from purely purely equilibrium considerations. We have not used any other tool for these derivations.
Example 5.9 A built-up beam Consider the built up beam shown in Fig. 5.36. It is held together with 10 mm bolts as shown. If the maximum shear force acting on the beam is 50 kN and the maximum shear stress a bolt can resist is 80 MPa, determine the maximum spacing s between the bolts.
This backward force is balanced by the shear force acting on the front face (of area t.dx) of this this element. If η xz is the average shear stress on this area
Solution:
.
To externalize the shear forces in the bolts, we consider the FBD of an element of length dx as shown in Fig. 5.37. Because of the presence presence of the shear force, the bending moments, and therefore, the bending stresses on the two opposite faces of this element are different. The difference in the axial forces because of this difference results in a net force on the element due to the bending stresses. This force is resisted by the shear action F s,b bolts. But there s,b in the bolts. Fig. 5.37 are two sides to this element. The unbalanced force is, therefore, resisted by 2 F s,b s,b.
From this we obtain
(5.16)
This is same as Eq. 5.15 with section width b replaced with t, the thickness of the flange. It is convenient to use the concept of shear flow q first introduced in Sec. 3.7 as write
∫
. Thus, both for for flange as well as web, we can
(5.17)
Fig. 5.35 shows the shear flow distribution for an Ibeam subject to a negative shear force V . If the shear force was positive, all the arrows would have been reversed. The shear flow in the flanges starts at zero at the
The unbalanced force is determined by Eq. 5.14 as
Each bolt resists this unbalanced load for a length s of the beam, where s is the distance between the bolts (i.e., the pitch).
.
Fig. 5.36 A built-up beam 72
The stress distributions at at the junctions are quite complicated. complicated. It is for this reason that the commercially available rolled-sections are provided with generous fillets to avoid stress concentration.
87
The value of I zz is evaluated by breaking up the beam section into a difference of two areas: a rectangle 140 mm wide, 160 mm deep, from which is subtracted a rectangle 120 mm wide, 140 mm deep. The value I zz is of I
The maximum value of the shear flow can be determined from Eq. 5.17 as VQ/I zz. The Q here is the first moment about neutral axis of the area of I zz is found by breaking the upper flange, which is ( bt )h/ 2. The value of I up the area into three parts, determining the I of each about its own centroidal axis, and then transferring it to the neutral axis using the parallel-axes theorem:
, -
The first moment of the area of the face of the FBD about the NA is obtained by multiplying the area (10 mm × 50 mm) by the distance of its ‒ 5 3 centroid from the neutrl xis xis (80 mm ‒ 5 mm). Thus, Q = 3.75×10 m .
The maximum value of q is then Vbht/2I zz. The total horizontal shear force F 1 in the flange is the area under the shear flow triangle, which is 2 ½qmax.b, or F 1 = Vhb t/ 4 I zz, pointing to left. The lower flange flange has the same value of shear force, except that it is pointed to the the right. The shear force in the web F 2 is equal the sectional shear force V.
The maximum shear stress in a bolt is prescribed as 80 MPa. Therefore, maximum value of F s,b s,b is
. /
. Using this
The resultant force on the section is now V , but the pair of forces F 1 produces a couple. Fig. 5.38(c) shows a statically equivalent equivalent force system in which the force V is displaced to a point O, a distance e to the left, with , or . The point O is termed as the shear centre of the section. As is shown in Fig. 5.39 5.39 the channel bends without without twisting only when the applied load passes through the s hear centre.
value in the equation above a long with V = 50 kN, we get
.
( )
This gives a value of s as 0.136 m. Thus, the spacing between the bolts bolts should be less than 13.6 cm.
5.8 Plastic deformations in beams Consider a beam in pure bending which is loaded beyond the onset of yielding. Let the beam be made of a perfectly elastic-perfectly plastic material whose behaviour is as depicted in Fig. Fig. 5.40. The yield stress of the material is ζ Y and the Y, corresponding strain is εY . As the bending moment applied to the beam Fig. 5.40 increases the magnitude of the strain at any point increases linearly with the bending moment. The strain for any applied moment varies linearly with y according to Eq. 5.1:
5.7 Shear centre Let us next consider beams with sections that have symmetry only about one axis, like a channel section shown in (a) (b) (c) Fig. 5.38. Let the the shear force on the section be V. Fig. 5.38 (a) shear flow distribution; (b) resultant forces in the three legs of the section; (c) point of We can easily find the action of the statically-equivalent shear force shear flow distribution along the three legs of the section. These are shown in Fig. 5.38a. The shear flow in the flanges increases from 0 at the tips to the maximum at the junction with the web. (a)
(b)
Fig. 5.39 (a) The channel channel twists while bending when the load does not pass
88
(5.1)
And as long as the strain is less than the yield strength value of εY , the bending stress variation is given by Eε xx, where E is the elastic modulus of the material. Under these (a) (b) (c) conditions, the stress Fig. 5.41 distribution is as shown in Fig. 5.41a. The value of ρ is found by considerations of equilibrium, so that the stresses are given by Eq. 5.5 as
The value of yY is 0 when the plastic region spans the whole depth of the beam. The beam in this condition is resisting the maximum possible moment it can resist. Thus, the maximum bending moment M max max that the beam can support is obtained by setting yY = 0 in Eq.5.19:
The value of ρ is easily determined in terms of yY by noting that the strain at yY Fig. 5.42 Variation of bending
.
is ‒ Y , so that
This is valid as long as εmax < εY , and consequently, ζ max max < ζ Y Y. The Value of bending moment M Y Y for which the maximum stress reaches the yield
, or
.
moment with curvature in elasticplastic deformation of a rectangular-section beam
, we get . Using this inin Eq. 5.19, we get [ ./], for ρ < ρ (5.21)
is given by
height of the section.
For a rectangular b×h beam, this gives
Y .
(5.18)
Fig. 5.41 shows the variation of bending moment with the radius of curvature of the beam for a rectangular-section beam. The maximum bending moment approaches a value 1.5 times M Y Y asymptotically. The ratio of the maximum bending moment to the yield bending moment changes with the section shape. For circular sections, this ratio is 1.7, and for thin-walled circular tubes it is about about 1.3. For typical commercially commercially available I-beams, this ratio varies between 1.1 and 1.2.
For values of bending moment larger than M Y Y, the material starts deforming plastically plastically at the either end. Though the strain distribution distribution is still linear, the bending stress in no longer linear with y. The value of the y less than h /2 as shown in Fig. stress acquires the value ζ Y Y at a value of y 5.41b and remains constant thereafter. If the radius of curvature is ρ for this case, then , and till its value is ζ Y Y, after which it remains constant at ζ Y The value of y at which ζ first Y. becomes ζ Y . We denote by yY . Thus, the beam is elastic Y is
Combining this with the fact that the radius of curvature ρY at the point where the plastic deformation just starts
, where h is the strength value is obtained from Eq. 5.6 as
(5.20)
to + y , and plastic outside this range. The value of ρ ρ
5.9 Strain energy in bending
within the rnge ‒ yY Y is as yet undetermined. We evaluate the bending moment by integrating integrating ζdA over the whole area.
The expression for strain energy in a structure subject to arbitrary stresses is given by Eq. 2.17 as
∫ ( )
∫ . / ∫ ∫ ∫ ∫ [ . /] [ . /]
(2.17)
where the integration is carried over over the entire volume of the beam. For the case of a beam in pure bending (no shear force and bending moment constant over the entire length), there is only one stress, ζ xx given by Eq.
or,
(5.19)
5.5 as 89
. The strain energy for such a beam is, therefore therefore
(5.22)
∫ ∫ ./ .∬ / ∫ . /
for 0 < x < L/ 2, 2,
for L/ 2 < x < L
For a uniform beam subject to a constant bending moment, this equation reduces to
And
.
(a)
for 0 < x < L/ 2 , for L/ 2 < x < L
(b)
In a more general case when a beam is also subjected to transverse shear, Eq. 2.17 gives the value of strain energy as:
The contribution of bending stresses to the total energy is given by
∫ ( ), or ∫ . /
∫
, where A is the cross-sectional area of the beam.
M ( x x), and replacing the value of M M ( x x) using Eq. Expressing ζ xx in terms of M b above, we get
(5.23)
∫ ∫ ∫
It was shown in Example 5.7 that the ratio of the maximum shear stress to the maximum tensile stress in the case of a uniformly loaded cantilever beam was . It can be shown shown that the same holds true true to almost all beams and loadings. This ratio is small for typical slender beams. Since the stresses are squared squared in Eq. 5.23, the the contribution of shear stresses to the strain energy is quite small compared to that of the bending (tensile) stresses and can, therefore, be safely neglected.
,
And expressing I zz in terms of b and h, we get
It can be verified that this has the dimensions of energy. energy. Please note that we have not included the energy due to shear which is an order of magnitude smaller for slender beams.
Example 5.10 Strain energy of a simply-supported simply-supported beam Consider the simply-supported beam of length L loaded at the midpoint as shown shown in Fig. Fig. 5.43. Find the strain energy of the beam if the section of the beam is rectangular with depth h and width b.
Summary
Solution: The reader can verify that the SFD and the BMD of the beam are as Fig. 5.43 drawn in the figure. Since we need to integrate the squares of shear stresses and bending stresses over the length of the beam, let us express SF and BM as functions of the longitudinal longitudinal variable x.
It can be seen that we can write: 90
A beam is a slender member which carries transverse loads or moments which tend to bend it. A beam resists bending moments by developing axial o stresses. These stresses arise because because curving of a beam necessarily implies contraction of material on one side and elongation on the other side. o There exists a plane within the beam which undergoes neither contraction nor nor extension. This is termed as the neutral plane. The trace of the neutral plane within a section of the beam is termed as the neutral axis. To obtain a relation between the curvature of the beam, the bending stresses and the moment causing the bending we adopt the following strategy:
We first use the geometry of bending in the case of pure bending to relate strains to the distance from the neutral plane. o We convert the strains to stresses using the elastic modulus. We next use the equilibrium equilibrium conditions. conditions. The balance of o axial forces gives us the location of the neutral axis as passing through through the centroid centroid of the the section area. area. The balance of moments then relates the curvature in the beam to the bending moment acting on it. Eq. 5.6 gives the relation between the moment M b, the radius of curvature ρ, the strain ε xx, and the bending stress ζ xx. The bending stresses vary linearly with distance from the neutral axis, zero on the axis axis and increasing as we move move away. There is compression on one side and tension on the other side of the neutral axis. We introduce the concept of bending rigidity as the curvature produced per unit bending bending moment. The bending rigidity rigidity for a uniform beam in pure bending is seen to be the product of the elastic modulus E and the second moment I zz of the crosssectional area about a transverse neutral (centroidal) axis. The rigidity of a beam increases as the third power of the depth of a beam. It was shown through some solved examples that for the beam sections of the same areas the beam with more material away from the centroid is more rigid. The method for handling of composite beams was developed following the the same process as for simple beams. beams. It was shown that strains in composite beams are determined by geometric considerations alone as in the case of simple beams, and are continuous across different different materials. The stresses, however, are not continuous. It was shown that the location location of the neutral axis is obtained by a relation that weights the areas of different materials with their elastic elastic moduli (Eq. 5.10). The flexural relation is the same as that for a simple beam except that the bending rigidity is now replaced by the sum of the bending rigidity of the constituent constituent areas. The process was explained using the example of a steel-reinforced concrete beam which exploits beautifully the differing capabilities capabilities of the two materials. o
The presence of shear forces causes the variation of bending moments along the length of a beam resulting in an unbalance of axial forces. This gives rise rise to shear stresses stresses η zx or η xz. The shear stresses are given by Eq. 5.15 as , where Q is the first moment of the area of the section above the location y, and b is the width of the section at y. The shear stress is the maximum at the neutral axis and decreases towards either end of the section. The shear stresses in slender beams are an order of magnitude smaller than the bending stresses. It is convenient to use the concept of shear flow q as
∫
. Thus, both for flange flange as well as web, we can write . The concept of shear centre was introduced as a point within the beam section through which the external load must pass for the beam to bend without twisting. twisting. It is the point where the single shear force which is statically equivalent to the sectional shear distribution must be applied. The strain energy in slender beams is dominated by the energy due to bending stresses and is given by
91
∫ . /
.
92
Example B.1 Centroid of a T-section Consider the T-section shown in Fig. B.2. Find the location of the c entroid.
Solution: The x location of the centroid is immediately obtained by using symmetry. Thus, xc = 2.5 units. The y-coordinate of the centroid can be obtained by using Eq. B.3.
Appendix B Properties of Areas Areas
We introduce here a very convenient Fig. A.2 T-section method for determining the centroid of areas that are made up of simple areas. Thus, we divide the T-section in to two rectangular rectangular areas, a 1×4 unit area labelled A, and a 4×1area labelled B. We know that centroid of area A is located at its mid-point, i.e., at y = 4.5 units. The centroid of area B is located at its mid-point, i.e., at y = 2.0 units. The location of the centroid of the composite area is given by
B.1 First moments of area and centroid Consider an area as shown in Fig. B.1. Two first moments of areas are defined: one about the x- axis, and the other about the y- axis.
∫
, and
∫
(B.1)
(∑ ) ∑
The values of Q x and Q y depend on the location of the origin O. The centroid of the area is defined as the location of the origin such that Q x and Q y vanish. We can use this property to determine the location of the centroid. Let the coordinate of the origin in the given coordinate system be xc and yc. We calculate the first moment moment of area about this point. point. The distances y and x in Eq. B.1 are the replaced by y – yc) and ( x x – xc), respectively: ( y
∫ ∫ ∫ ∫ ∫
Using this formula, we get
units
The location of the centroid centroid is shown in the figure. located nearer the top since the area is top heavy.
The centroid is
B.2 Second moments of area
Fig. A.1
A second moment of area is a moment in which two lengths are used. We 73 define three second moments :
, or
, or
And similarly,
(B.4)
73
The nomenclature of the second moments has two indices, each indicating the axis from which which the distance distance is taken. Thus, I zz uses the distance y from the z axis twice, I yy uses the distance z from the y axis twice, and I yz uses the distance y from the z axis as well as the distance z from the y axis. This last is also called a cross moment of area . The polar moment of area used in Chapter 3 is also a second moment where the distance of the area is measured from the polar axis, i.e., the axis normal to the area. It will be labeled I xx in the convention used here:
(B.2) (B.3)
93
∫ ∫ ∫ ;
, and
Solution:
(B.5)
4
We can, using symmetry, see that the centroid will be at the centre of the section.
4
The dimensions of these second moments are L , and its SI units are m .
∫ ∫ ∫
Example B.2 Second moment of a rectangular section Determine the second moment of area of the rectangular section shown in Fig. B.3 about its centroidal axis.
Recognizing that the first integral is the I xx of the outer square (= 3 3 a.a /12), and the second integral is the I xx of the inner square (= b.b /12), we get the I xx of the section as .
Solution: The centroid of this symmetric section obviously lies at the middle middle of the area. We take the coordinate axes with the centroid as the origin. Let us take a rectangular elemental area of dimensions δy and δz . Then, by Eq. B.5:
∫ ∫ ∫
This is a general method of determining the moments of composite sections.
B.3 Parallel axes theorem Fig. A.3 A rectangular section
We now introduce a very important property of the second moments of area which is very handy in calculations of the moments. Refer to Fig. Fig. B.6. Consider Consider First a coordinate system x-y with the origin O and x-axis passing through the centroid. The moment in this coordinate system is given by
Carrying out the indicated integrations, we get: (B.6)
Similarly,
.
We next calculate these moments about the y-axis that coincides with the bottom plane of the section.
∫
Fig. A.4
∫ ∫ ∫
four times the value when the x-axis passes through the centroid. Clearly, the value depends on the location of the x-axis. We shall show below that that the value with the x-axis passing through the centroid is the minimum of all possible values.
∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫
or,
Since y is measured from the centroidal axes, seen as I xx,
Example B.3 Second moment of a hollow square section Determine the second moment of area of the hollow square section shown in Fig. B.5 about its centroidal axis.
Fig. A.6
Next consider an arbitrary coordinate system x’ - y’ y’ with the x’ -axis located a distance d away from the centroidal axis. The second moment in this new system is
hence zero, and
is immediately
is the first moment of area about a centroidal axis and is nothing but the total area A. This gives:
(B.7)
Fig. A.5 Hollow square section
94
This is known as the parallel axes theorem . The second moment about an axis a distance d away from the centroid is equal to the moment about the centroidal axes plus area times the square of d.
Example B.5 Second moment of an angle section about the centroidal axis Consider the angle section shown in Fig. B.8. Determine its second moment about the centroidal axis.
One can immediately see from this theorem that the second moment about the centroidal axis is the minimum. This theorem provides provides a very convenient tool to obtain moment of inertia of a composite area as is illustrated below.
Solution: We first determine the location of the centroid of the the section. We work with the division of the section in two rectangles A 2 and B as shown. The area of A is 400 mm and its centroid is at (5 mm, 30 mm) and 2 B is 500 mm and its centroid is the area of B at (25 mm, 5 mm). mm). The coordinates of the centroid are:
Example B.4 Second moment of a T-section about the centroidal axis Consider the T-section T-section of Example B.1. Determine its second moment about the centroidal axis.
Solution: The centroid of the section was determined in Example B.1 to be located on the centre line a distance 3.25 units up from the bottom bottom of the T. The origin of of the axes is now taken at this location as shown.
and
We split up the area of the section into two rectangles A and B as before (Fig. B.7).
∑∑ ∑∑
Fig. A.8 An angle section
mm, mm.
The location of the centroid has been marked on the figure.
Here, too, we calculate the second moment using the parallel axis theorem. We first calculate the moments of the two areas about their own centroidal axis, and the transfer them to the centroidal axis of the composite section:
We can calculate the second moment of Fig. A.7 A about its own centroidal axes (at area of A 3 a distance of 4.5 units from the bottom) using Eq. B.6 as (5×1 )/12 = 0.417. We then use the parallel axis theorem (Eq. B.7) to shift the axis to 3.25 units above the bottom bottom (from 4.5 units above the bottom. Thus, the shift distance d is 4.5 – 3.25 = 1.25, and the value of becomes 2 0.417 + (Area = 5×1)×(1.25) = 8.23 units. Similarly, the second moment of area B about its own centroid (which is 2 units from the 3 bottom) is 1×4 /12 = 5.33 units. When we transfer it to the centroid of the the whole section which is a distance d (= 3.25-2.0 = 1.25 units) away, it 2 becomes (Area = 5×1)×(1.25) = 13.14 units. Thus, the net moment is units.
, 0 1 1 0
4
mm .
95
∫ ∫ ∫ ∫
Thus, the second moment of this angle section about the centroidal axis ÷7 4 has been determined as 2.99×10 m .
the polar moment of area
. But
. Thus,
Table A.1 Properties of some areas Geometry
Rectangle
Area
Vertical location of centroid
Second moment of area about centroidal axis
bh
h/ 2
bh /12
h/ 3
3
. Therefore,
(B.8)
This is the perpendicular axes theorem which states that the polar moment of inertia of a section is the sum of the second moment of the area about two perpendicular axes within the plane of the area.
Example B.6 Second moment of a circular section Fig. 3 A.8
Determine the second moment of a circular section.
Solution: Isosceles triangle
bh/2
Consider the circular section of Fig. B.9. We
bh /36
∫
need to determine . Direct determination of this integral is not easy. Therefore, we use the perpendicular axes theorem (Eq. B.8): Circle
πD2 /4
πD4 /64
D/ 2
Fig. A.9 A circular section
The symmetry of the circular section gives us ensures that the moments and are equal. Thus, . The polar moment had been
I-beam
[2k 1+(1‒2k 1)] bh
h/ 2
evaluated in Sec. 3.3 as
, -
B.4 Perpendicular axes theorem Consider an area as shown in Fig. B.8. Let us consider a right-handed triad of coordinate axes with x- and y- axes within the plane of the section and the z-axis pointing outwards outwards from the paper. Using the definitions definitions of
96
(Eq. 3.6). Therefore,
Index area method, 64 area, cross moment of, 93 area, first moment, 84, 93 area, polar moment, 43 area, second moment, 43, 93 axis, neutral, 73, 91 beam, 56 beam supports, idealized, 58 beam, built-up, 87 beam, composite, 78 beam, plastic deformations in, 89 beam, sandwich, 80 beam, shear stresses, 82 beam, simply-supported, 58 beams, shear flow, 87 bearing stress, 5 bending moment, 7, 56 bending rigidity, 74, 91 bending strain, 73 bending stress, 73 bending, pure, 71 bending, strain energy, 90 built-in support, 58 built-up beam, 87 cable in a sleeve, 45 Castigliano theorem, 33, 40 centroid, 74, 93 coefficient of thermal expansion, 27 coil spring, 53 complementary planes, 12 composite beam, 78 composite shaft, 49 concrete, pre-stressed, 32 concrete, reinforced, 81 cross moment of area, 93 deformation, plastic, 29
distributed load, 59 elastic limit, 29 elastic modulus, 5, 17 electroplating, residual stresses in, 27 energy methods, 33 energy, strain, 33 first moment of area, 84, 93 gauge length, 29 generalized Hooke law, 24 Hooke law, generalized, 24 Hooke, Robert, 5 idealized beam supports, 58 idealized stress-strain curves, 30 key, 10 keyway, 10 limit analysis, 52 limit load, 52 linear elastic range, 29 load, distributed, 59 load, point, 58 loading density, 59 method of sections, 61 moment, bending, 7, 56 moment, twisting, 6 necking, 30 neutral axis, 73, 91 neutral plane, 72, 91 normal stress, 11 parallel axes theorem, 95 Pascal, 7 permanent set, 29 perpendicular axes theorem, 96 pinned support, 58 plane, neutral, 72, 91 planes, complementary, 12 plastic deformation, 29
plastic deformation in torsion, 51 plastic deformations in beams, 89 point load, 58 Poisson ratio, 23 polar moment of area, 43 pre-stressing, 32 proportional limit, 29 radius of curvature, 73 rectangular section, second moment of, 94 reinforced concrete beam, 81 right-hand thumb rule, 44 rigidity, bending, 74, 91 rigidity, torsional, 43 roller support, 58 sandwich beam, 80 second moment of area, 43, 93 second moment of rectangular section, 94 shaft, composite, 49 shaft, statically indeterminate, 47, 54 shaft, tapered, 45 shear centre, 88 shear flow, 50 shear flow, beams, 87 shear force, 56 shear modulus, 26 shear strain, 25 shear strain in torsion, 42 shear stress, 6 shear stress in torsion, 42 shear stresses in beams, 82 shrink fit, 28 sign convention for shear force and bending moment, 57 97
simply-supported beam, 58 spring, kinky, 22 St. Venant principle, 8 statically equivalent loads, 8 statically indeterminate problems, 20, 37 statically indeterminate shaft, 47, 54 strain, 5 strain energy, 33 strain energy, bending, 90 strain energy, elastic body, 38 strain energy, torsion, 53 strain hardening, 30 strain, bending, 73 strain, shear, 25 strength, yield, 29 stress, 5 stress at a point, 7 stress on oblique planes, 11 stress, bearing, 5 stress, bending, 73 stress, double-index notation, 12 stress, shear, 6 stress, sign convention, 13 stresses, pressure vessel, 14 support, built-in, 58 support, hinged, 9 support, pinned, 58 support, roller, 9 symmetry arguments, 15 tensile test, 28 thermal expansion, coefficient of, 27 thermal strains, 27 thermal stresses, 27 torsion, 41 torsion bar, 41
torsion moment diagram, 44 torsion, plastic deformation, 51 torsion, strain energy, 53 torsional rigidity, 43
torsional strain, 42 torsional stress, 42 twisting moment, 6 twisting moment diagram, 44
two-force member, 9, 10 ultimate stress, 30 unit force method, 35 vibration isolator, 26
98
yield point, 29 yield strength, 29
