Strength of Materials
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MULTIVECTOR Review and Training Center Rm. 867, Ground Floor, Isabel Bldg. F. Cayco comer Espafia Sts. Sampaloc, Manila Tel. No. 7317423
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MUL T.IVECTOR REVIEW AND TRAINING CENTER STRENGTH CHt~TERIALS STRENGTH OF MATERIALS - is a measure of its ability to resist the application ofthe load required during the service of the structure of the machine without fracture, collapse or undue distortion. Stress- is the mechanical force on energy that causes or produces deformation or fracture in the material. -defined as resistance to external forces and it is measured in terms of the force exerted per unit of area. 1- Simple Stresses: I.
Axial Stress: a. h.
Tensile stress Compressive stress
Formula: p
S=A
where: A .l P S =stress P = load or total external load . A =stressed area 2.
Shearing Stress:
Formula: Ss ==
~~
projected area
where: As// Ps . 3.
Bearing Stress
Formula:
p
Sb=~ where Ab is the projected area
II Factor of Safety (FS)- is the ratio of ultimate stress to allowable stress.
_ Greatest .stress of'\ FS =·Ultimate stress ~ material can withstand Allowable stress , without rapture Note FS > I Thin-Walled Pressure Vessel:
I Cylindrical Shape
D
'Workin~·stress or safe working stress
MULTIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATEl~lALS F
a) Circumferential Stress (Sc)- stress at the side.
p
p
A
A
F = Bursting force
P = Resisting force
L:Fv = 0 2P=F 2 Astress · .Sc = pA 2 (LX t) (Sc) = p (D XL)
_.212_
Sc- 2t a.
Longitudinal Stress (SL)- stress at the end:
~,..
p
,.
1\
'
\
II
__ /
II
I
'-..:......---_.... ~--------' ',_f'~.:-: t /
. . . . . . . . . . . . . . . . . . . . ..
p = C0h p = hydrostatic pressure or gauge pressure in the tank Inner Diameter D = diameter · t = thickness C0 =density of the fluid h = elevation head where:
Asu:ess · SL = P A n D t SL = p ~ D 2
.
4
SL= pD 4t Note:
~
I\ \
l::FH = 0 P=F
-~---------------
t .. , ,/,/-r'::,~",~,.-------- ~ 1 1 D I I :-.:: \ i i I F --'f.l',-.1'+---.t--
SL = ~ Sc
II Spherical Shape:
n
?
F= p-D4
L:F = 0 P=F nDtS = p~D 2
4
s = J2.Q_ 4t
P = n Dt S
MULTIVECTOR REVIEVV ..1!\.ND TRAINING CENTER
STRENGTH OF .MATERIALS Rotating Steel Ring
F= mo/r
IFH=O
2P"" F 2AS =
where:
molr
F = bursting force/ centrifugal force P =resisting 'force
m =.mass of one-half of the ring A= cross-sectional area of the ring V = volume of one-half of the ring p = mass density
r=l!£_ 11:
l'c "'' mean radius co = angular velocity; rad/sec v = peripheral velocity of the ring
Hool
Axial Deformation: I.
S'~Ee
2.
s"' 7:
p
where:
II
L - original length 8 - change in length P - load or force exerted A - area stressed
S- stress. e- strain . p E- modulus of elasticity or Young's modulus
Shearing Deformation:
tanY=_h L
However, since the angle 'Y is usually very smalf ----7-Ps
tan 'Y "''Y in radian l-Ienee
y
__§s_ yL
---/ I
or
bs
= 'Y
L
I
I
I
~-
:S..j
I
y
r-./ I /_
I
I
/T L
~
MULTIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATERIALS Assuming Hooke's Law to apply to shear; is or From
G
85 = y L = .&_L
= (Ps/A 5 ) L
G 8 s where
y=~
=
G
Ps L As G
y - shearing strain -the angular change between two perpendicular faces of a differential element
Spring in Series
w
w (Parallel)
(Series)
Thermal Deformation (8T)
where: a = coefficient of linear expansion
L = original length L1 T = change in temperature
MUL TIVECTOil. REVIEW AND TRAINING CENTER ' STRENGTH OF MATERIALS Torsion: T
t.
Torsional shearing stress (S 5 ):
Ss
=
-.Jf- {--
general fommla
For maximum Ss; p =, R =
2o.
Ss = TR ' J Where: p =distance from theN. A.(Center line for round specimen) J = polar moment of inertia T =torque R =radius 2.
Angle of Twist (Angular Deformation): TL
8=JG
e = angle of twist in radian
where:
L =length . G = modulus of rigidity 3.
Shaftings: a. Solid Circular Shaft:
Ss
=
~~7) 32
S = 16T s
b.
rr03
Hollow Circular Shaft: J =lL (04- d4)
32
Ss
T (D/2) _ d4)
_!I_ ( 0 4
32
l6T 0
Ss
rr (04 _ d4)
MULTIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATERIALS 4.
Power Transmitted by the Shaft (P): P = 2nNT Where: N -speed of rotation of the shaft in rev per unit time T- torque Note:
1 hp. = 550 ft-lb sec
ft-lb 33,000 - . - = 746 watts mm
Flanged Bolt Coupling:
T=~PR
T = nPR = n AbolrS,R T = R Abolt n S,
where
T = torque capacity S, =allowable shearing stress in each bolt Abolt = area of each bolt R = bolt circle radius n =number of bolts in the bolt circle
Coupling with Two Concentric Bolt Circle:
T=L:PR T = n 1 P 1R 1 + n2 P 2R2 , where y1and y2 are shearing strains or shear deformation
I
I
e
MUL TIVECTOR REVIEW AND TRAIN{NG CENTER STRENGTH OF MATERIALS
.
.
Using Hooke's Law for shear; G = Ss
y
~=~
we have, r·
G,R, G2R2 _fJ!.&=~ G 1R1 G2Rz
If the bolts on the two circles have the same area, A1 = A2; and ifthe bolts are made of the same material, G 1 = G2, then it reduces to
J~elical
Spring:
p
where: d =.diameter of steel wire R = mean radius of spring
Spring DcflcctiorL(o)
0
64 PR 3n Gd 4
Maximum Stress (S 111 . .): Approxi1i1ate Formula:
l2_PR.[ I +L 1t
ce
4R
16 P~ [I + _Qj_ ] 7td
111
A.M. Wahl Formula: · 16PR [4m-I + 0.615] Smax=~ 4m-4 m where
m = spring index D
m= d
=
2R
d
. Note: A.M. Wahl Formula is more precise formula usually used for heavy springs which are sharply curved where m is not so large while formula I is usually used for light springs where the ratio m is large. ·
MULTIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATERIALS ImpactStress
.~
Smax = Sst [ 1 + ~ 1·
-t-
Ost ]
If
h
Where: Smax- impact stress . Ss 1 - static stress h- height of fall Ost - static deformation
-*
:CY
- - - - .., ___',j(
:_--- ~
8
Note: Impact factor=--;: USt
Flexural Stress _Me SrINA
z=
INA
c
where
M - maximum bending moment c - distance from the center to the farthest fiber. INA- moment of inertia with respect to neutral axis z- section modulus
I
I
(liVre~
~(#)")
-
\-
- d,t
~~=~=;/~
MULTIVECTOR REVIEVII AND TRAINING CENTER STRENGTH OF MATERIALS
REE- Apr. 2006 1. What is stress? A force per unit length B. mass per unit acceleration
Q. force per unit area D. force per unit volume
REE- Oct. 2000 I May 2009 2. Find the required diameter of a steel member if the tensile design load is 7,000 pounds. Assume a safety factor of 5 based on an ultimate strength of 60,000 lbf/sq. in. A 5.406 em ji. 0.8618 in C. 4.695 mm D. 3.2367 m REE - Sep. 2005 3. A 1000-kg homogeneous bar AB 2 m long is suspended horizontally from two vertical cables AC and BD, each 1.8 m long and connected at both ends, with cross-sectional· area 400 mm. sq . . Determine the magnitude of the largest additional force acting downward which can be applied to the bar. The stresses.in the cables AC and BD are limited to 100 MPa and 50 MPa, respectively. fl. 50.2 kN B. 55.0 kN C. 59.0 kN D. 67.0 kN REE -Oct. 1996 I Oct. 2000 I Sep. 2003 l Apr. 2005 4. Determine the outside diameter of a hollow steel tube that will carry a tensile load of 500 kN at a stress of 140 MN/m. sq. Assume the wall thickness to be one-tenth of the outside diameter. A.111.3mm B.109.7mm C.113.7mm D.112Amm REE -. Sep. 2004 5. What is the force to punch a 0.75 inch circular hole through a piece of 5/8 inch thick steel plate (ultimate shear strength of 42 ksi)? A 154 kip B. 732 kip C. 43 kip }2: 61.85 kjp 6. A hole is to be punched out to a plate having an ultimate shearing stress of 310 MPa. If tt1e compressive stress in the plate is limited to 420 MPa. What is the maximum plate thickness in which a hole of 75 mm diameter can be punched? A 25.4 mm B. 18.5 mm C. 32.5 mm D. 27.8 mm REE - Apr. 2006 7. A thin walled pressurized vessel consists of a right circular cylinder with flat ends. Midway between the ends, the stress is greatest in what direction? A. Longitudinal B. Circumferential C. Radial D .. At an angle of 30 degrees to the longitudinal and circumferential direction 8. In designing a cylindrical pressure tank 3 feet in diameter, a factor of safety of 2.5 is used. The cylinder is made of steel (yield stress 30 ksi) and will contain pressures up to 1000 psi. What is the required wall thickness, t, based on circumferential stress considerations? A. 0.75 in B. 1.50 in C. 3 in D. 3.75 in
=
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REE - Oct. 1996 I Oct. 1999 / Sep. 2004 9. Find the limiting peripheral velocity of a rotating steel ring if the allowable stress is 140 MN/sq. m. and the mass density of the steel is 7850 kg/cu. m. A. 130.62 m/s B. 129.58 m/s C. 142.37 m/s D. 133.55 m/s REE - Sep. 2003 I A.pr. 2007 10. A rotating steel ring has a mass density of 7,850 kg/cu. m. At what angular velocity will the stress reach 200 MN/sq. m. if the mean radius is 250 mm? A. 750.30 rad/sec B. 564.33 rad/sec C. 700.21 rad/sec D. 638.47 rad/sec
~.
MJlL TIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATERIALS REE -Apr. 2002 11. The simple mathematical statement of the relationship between elastic stress and strain : stress is .. ·proportional to strain. What is this law? A Boyle's law B. Hooke's law C. Charle's law D. Gas law. REE - Apr. 2002 12. For normal stress, what is the constant of proportionality? A Modulus of plastiCity' £ Modulus of elasticity B. Modulus of cavity . D. Modulus of varsity . REE - Apr. 2005 13. What is the decrease in height of an 8-inch round x 16-inch high concrete cylinder (E = 2.5 x 10 to the positive 61h powerlb/ sq. inch) when the unitdeforrilation is 0.0012? · A 392m B. 0.0192 in C. 355 em · D. 891 mtn REE - Apr. 2001 I Sep. 2009 14. What is the stress in an 8-inch round x 16-inch high concrete cylinder (E = 2.5 x 10 to the positive 61h power lb/sq. inch) when the unit deformation is 0,0012? A 500 lbf/sq. in. (psi) ·C. 8,100 lbf/sq. in. (psi) B. 210 lbf/sq. in. (psi) D. 3,000 lbf/sq. in. (psi)
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REE .,... May 2010 15. Find the elongation of the aluminum (in inch) specimen 100 inch long when loaded at each point. E = 10 x 10 raised to 6 psi and yield stress = 37 ksi. Neglect the weight of the bar. A. 0.65 B. 0.37 C. 0.43 D. 0.25 REE - Apr. 2002 I Apr. 2004 I Sep. 2005 16. A steel rod having a cross-sectional area. of 300 sq. mm. and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/cu. m. and E = 200 x 10 to the positive 3 MN/sq. m., find the total elongation of the rod. A. 52.48 mm B. 60 mm C. 54.33 mm D. 56 mm REE -Apr. 2007 17. A reinforced concrete column 250 mm. in diameter is designed to carry axial compressive load of 400 kN. Using allow~ble stresses of concrete 6 MPa and steel at 120 MPa, find the required area of stresses of reinforcing steel. Assume Ec = 14 GPa andEs= 200 GPa. A. 1720 sq. mm. B. 1600 sq. mm. C. 1320 sq. rnm. D. 1440 sq. mm. REE - Sep. 2010 18. A 10 mm diameter round bar is made of aluminum alloy 7075-T6. An axial force stretches the bar resulting in a decrease in its diameter by 0.016 mm. Find the magnitude of the axial load, in kN. A. 27.4 B. 28.3 C. 25.6 D. 24.7 REE - Apr. 1996 19. A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 em and a cross-sectional area of 15 em. sq~ The shear modulus for this rubber is 2 MPa. If a sideways force of 300 N is applied to the motor, how,farwill it move sideways? A. 0.124 em B. 0.075 em C. 0.102 em D. 0.092 em /
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20. A 20 N weight is supported by two springs in series. The first spring has a stiffness of 1 N/mm and the second spring has a stiffness of 2. 0 N/mril. The natural period of vibration is A. 0.55 sec B. 0.35 sec C. 1.15 sec D. 0.75 sec
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e Vl
MlJLTIVECTOR REVIEVV AND TRAINING CENTER STRENGTH OF
MATERIA~S
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•. .. .
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21. Determine the stresses set up in the rails of the tramway when the temperature rises from 0 to . 50 deg F. The rails are welded together so that there are no clearance and are laid so that they cannot expand. For steel; assume a = 6.5 x 10 to the negative 6 per deg F and E = 20 x 10 to the positive 10 N/m. sq. A 55 MPa B. 60 MPa C. 65 MPa D. 70 MPa
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REE - Apr. 2002 22. At 90 deg F, the stress in a steel roo is 2000 psi (C). What is the stress at 0 deg F? A. 15,550 psi (T) B. 53,900 psi (C) C. 68,210 psi (T) D. 27,430 psi (C)
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)
REE - Sep. 2001/ Sep 2002 I Sep. 2005 23. A solid steel shaft 5 m long is stressed to 60 MPa when twisted through 4 degrees. Using G = 83 GPa, compute the shaft diameter. A. 0.104m B.1.174cm C. 5.214mm D. 9.432in
,
REE- Apr. 20011 May 2008 24. A solid steel shaft 5 m long with diameter of 0.104 m is stressed to 60 MPa when twisted through 4 degrees. Determine the power that can be transmitted by the shaft at 20 rad/sec. A. 1.9 MyY B. 2.1 MW C. 1.6 MW D. 2.5 MW ~
f:
REE - Sep. 2003 25. A hollow bronze shaft of 75 mni outer diameter and 50 mm inner diameter is slipped over a solid steel shaft 50 mm in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. Find the maximum shearing stress developed in the bronze by end torque of 3 kN-m. For bronze G 35 GPa and for steel G 83 GPa. A. Te =28.5 MPa B. ta =84 Pa C. ta =10 kPa D. 'te =65.2 GPa
-
=
=
.
REE - Apr. 2006 I Sep. 2009 26. A flanged bolt coupling consists of eight steel 20 mm diameter bolt spaced evenly around a bolt circle 300 mm diameter. Determine the torque capacity of the coupling if the allowable shearing stress on the bolts is 40 MN/m. sq. A. 22.1 kN-m B. 20.5 kN-m C. 15.08 kN-m D. 18.4 kN-m
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REE- Sep. 2002 27. A flanged bolt coupling consists of six-1 0 mm diameter steel bolts evenly spaced around a bolt circle 300 mm in diameter and four-20 mm diameter aluminum bolts on a concentric bolt circle 200 mm diameter. What torque can be applied without exceeding a shearing stress of 60 MN/m. sq. in the steel or 40 MN/m. sq. in the aluminum? A. 9.75 N-cm B. 2.8 MN-mm C. 5.94 kN-m D. 4.5 GN-in REE- Apr. 2001 I Apr. 2003 I Apr. 2004 28. Determine the maximum shearing stress in a helical steel spring composed of 20 turns of 20 mm diameter wire on a mean radius of 80 mm when the spring is supporting a load of 2 kN? ·A. 140 MPa B. 115 MPa C. 121 MPa D. 160 MPa
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REE - Sep. 2004 29. A 14 foot long simplE~ beam is uniformly load with 200 pounds per foot over its entire length. If the beam is 3·.625 inches wide and 7.625 inches deep, what is the maximum bending stress? A. 6332 lbf/sq. in. (psi} C, 7974 lbf/sq. in. (psi) B. 1674 lbf/sq. in. (psi) D. 8205 lbf/sq. in. (psi)
MULTIVECTOR REVIEW AND TRAINING CENTER STRENGTH OF MATERIALS REE - Apr. 2001 I Apr. 2003 30. The beam shown in Figure 3 is 5 inches wide and 10 inches deep. What is the maximum bending .. stress? · A. 7,721.0 lbf/sq. in. (psi) .Q..2,888.61bf/sq. in. (psi) B. 5,500.3 lbf/sq. in. (psi) D. 3,420.9 lbf/sq. in. (psi) 2000 pounds
.
1300 pounds/foot
Figure 3
REE - Sep. 2001 . 31. A 4 inch wide, 8 inch deep wood beam is simply supported and loaded as shown in Figure 5. What is the maximum shearing stress? A. 449.1- lbf/sq. in. (psi) C. 778.7 lbf/sq. in. (psi) ~ 284.4 lbf/sq. in. (psi) D. 254.2 lbf/sq. in. (psi) 4000 pounds
3'
4'
5'
4333
6067
Figure 5
REE -Apr. 2002 32. What is the maximum stress in the offset link shown in Figure 7? Neglect stress concentration . · factors. A. 15, 676.6 lbf/sq. in. (psi) .Q.... 12, 666.7 lbf/sq. in. (psi) B. 20, 534.9 lbf/sq. in. (psi) . D. 50, 984.2 lbf/sq. in. (psi) I
'
· Load ,;, 4000 pounds (tension)
Figure 7
REE - Apr. 2002 I Sep. 2003 33. What uniform load will cause a simple beam 10 feet long to deflect 0.3 inch if it is supported (in addition to the end supports) by a spring at the beam's midpoint? The spring has a spring constant cif 30,000 pounds per inch. Assume the beam is steel, 10 inch deep, rectangular, and wijh a centroid a! moment of inertia of 100 inches 4 . ~5440 lbf/ft B. 2100 lbflft C. 8970 lbflft D. 4550 lbf/ft
