Principle of Statics.pdf

Hong Kong Kong Institute nstitute of Vocationa Vocationall Education Education (Tsing (Tsing Y i) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 1 – PRINCIPLES OF STATICS

PRINC PRI NCII PL PLES ES OF OF STATICS TI CS Statics is the study of how forces act and react on rigid bodies which are at rest rest or not not in in motion. otion. This his stud study y is is tthe he basis sis for the engi ngineering ring princi principl ple es, which guide the design of all structures, since before we can begin to design any structure we must first know the forces applied to it. Newton’s third law of motion states that: F or every acti action on ther there is alw always ays an equal equal and oppo opposi site te reacti action. The The study of st statics ics has many applica lication ions in daily life. An Any yone who has used used a ladder dder should should have have a dee deep appreci apprecia ation tion for for the laws that govern govern the stabi stabillity of the ladder. Hopefull ully the the reacti reactions ons gene generated rated by the ground surf su rfa ace and the wal wall su surf rfa ace are su suffficie cient to maintain equi quilibrium brium. I f these forces cannot be developed, an accident will likely occur.

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Hong Kong Kong Institute nstitute of Vocationa Vocationall Education Education (Tsing (Tsing Y i) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 1 – PRINCIPLES OF STATICS

EQUI EQUI L I BRIUM RI UM OF STRUCTURES STRUCTURES A structure is considered to be in equilibrium if it remains at rest when subje subject to to a system of forces orces and and mome oments. I f a struct structu ure is in in equ quiilibrium, then all its members and parts are also in equilibrium. For a structure to be in equilibrium, all the forces and moments (including support support reactions) ctions) acti acting ng on it it must balance each othe other. For a pla plane structure subject to forces in its own plane, the conditions for equilibrium can can be expresse xpressed by the foll ollowing owing equations of equilibrium:

∑ Fx = 0,

∑ Fy = 0, ∑ M z = 0

The The third ird equation ion above states that the sum of moments of all for forces about any point in the plane of the structure is zero.

∑ F = 0, F1 + (-F2) = 0 F1 = F2

∑ Fy = 0, F3 + (-F4) = 0 F3 = F4

F orces orces and and thei their resul ultants tants Forces orces are vectors, vectors, and becau because se of this this they they have have both ma magnitude gnitude and direct di rectiion. This his is is extremely im important portant to rea realize becaus cause e an analysis ysis of of a structure cannot cannot be comple pleted ted by sim simply ply knowi knowing ng the magnitude gnitude of forces without the direction of application.

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Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 1 – PRINCIPLES OF STATICS

Techniques of finding resultants In solving statics problems, often it is an advantageous to replace all the forces, which act on a body with a single force. This single force, called a resultant, must produce the same movement and effects that all the original forces would produce on the body.

• Vector addition using trigonometric functions • Resolution into rectangular components Vector addition using trigonometric functions The resultant of two vectors can be found by placing the two vectors (which act at the same point) in a tip-to-tail fashion and completing the triangle with a vector.

The resultant vector (c) is the sum of the original two vectors (a and b) and therefore replaces the original two vectors. By using Cosine Law, c2 =a2 +b2 –2*a*b*cos C The resultant of three or more vectors can be found by placing the vectors (which act at the same point) in a tip-to-tail fashion and completing the polygon.

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The polygon method requires finding an intermediate resultant, r’. This intermediate resultant is to be the sumof the two preceding vectors and will subsequently be added to the next vector to find another intermediate resultant. This process will repeat itself until thelast vector has to be added to the last intermediate resultant.

Finding the resultant of parallel force systems The resultant of a parallel force systemmust have the same net effect on the body as the parallel system, which it is replacing. That means the translational characteristics, as well as therotational characteristics, must be equal. 4/23


Example The following body is acted upon by a group of parallel forces. Find the resultant of the resultant.

Solution Considering the translational effects, the magnitude of the resultant can be found by simply summing the forces involved.

R =5 +8 –3 N =10 N This 10 N force has to act at a particular point that gives the resultant the same rotational characteristics as the real system. If point O is chosen as the reference point, the moments from the real system would be as follow: Take moment about point O, R*x = 5*3 + 8*7 – 3*11 x =3.8 m 5/23


Equilibrium of a structure A loaded building shown in the following figure is in equilibrium if it does not move as a rigid body. Rigid body movement can be either a translation (movement in a straight line) or a rotation or a combination of both. In this case we must have some physical means to tie up the building on to the ground in the first place, and through these ties the load can be transmitted from the structural to the ground. These ties are called restraints, or constraints or supports.

∑ Fx = 0,

∑ Fy = 0, ∑ M z = 0

Tie up a body or a structure A body in a plane has three degrees of freedom, which means that the body is free to move about in x-direction, in y-direction as well as rotation. If we want to maintain this body in equilibriumand prevent it from moving away under the loads, we must tie it up with some ties to the foundation.

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It is seen that we have used two ties to tie up the body of A. This body now cannot have translated displacements in both x and y directions. But it can still rotateabout A. In order to prevent it from rotation too, we used another tie to tie up the body.

In conclusion, at least three ties are necessary in order to tie up a body onto a foundation so that it cannot disappear or rotateunder the action of external loads, and it can be maintained in the conditions of equilibrium.

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Note that the above arrangements of the three ties or restraints cannot maintain the body, or the structure, in equilibrium.

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TY PES OF STRUCUTRES

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SUPPORTS Pin or Hinge Support

A pin or hinge support is represented by the symbol

HB

Prevented: Allowed:

VB

or

H

B

VB

Horizontal translation and vertical translation Rotation

Roller Support

A roller support is represented by the symbol or

VB

Prevented: Allowed:

VB

Vertical translation Horizontal translation and Rotation

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Fixed Support

A fixed support is represented by the symbol MB

HB

or VB

Prevented: Allowed:

HB

V M

B

B

Horizontal translation, Vertical translation and Rotation None

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Example 1 Beam ABCD has a pinned support at A and a roller support at C. It carries two concentrated loads of 20 kN each and a uniformly distributed load of 4 kN/m over the right hand half as shown. Determine the reactions. 20 kN 20 kN 4 kN/m A B 3m

C

D

1.5m 1.5m

3m

Solution: 20 kN 20 kN 4 kN/m A B

HA

VA 3m

∑X =0,

C 1.5m 1.5m

D

VC

3m

HA =0

Take moment about A, 20*3 + 20*9 + 4*(4.5)*(4.5 + 4.5/2) – V c*6 =0 ⇒V c =60.25 kN

∑ Y =0, 20 +20 +4*4.5 =V A +V C ⇒V A =-2.25 kN, (-ve sign indicates V A acts in opposite direction) ⇒V A =2.25 kN (↓) 12/23


Example 2 Find the support reactions for the simple beam shown. 5 40 kN

4

3

50 kN

A

D B

C 2.5m

5m

2.5m

Solution: 5 40 kN

40 kN

A

HA

50 kN 30 kN

B

4

3 D

C

VA

VD 5m

2.5m

2.5m

Resolve the 50 kN inclined external load into horizontal and vertical components as shown.

∑X =0,

HA =30 kN

Take moment about A, 40*5 + 40*7.5 – V D*10 =0 ⇒V D =50 kN

∑ Y =0, 40 +40 =V A +V D ⇒V A =30 kN 13/23


Example 3 Determine the truss reaction forces.

30 kN

30 kN 20 kN

A

m 5 B

5m

5m

5m

5m

Solution:

30 kN

30 kN

m 5

20 kN A HA

B V

∑X =0,

A

5m

5m

5m

5m

VB

HA =20 kN

Take moment about A, 30*10 +30*15 – 20*2.5 – V B*20 =0 ⇒V B =35 kN

∑ Y =0, 30 +30 =V A +V B ⇒V A =25 kN 14/23


Example 4 Determine the support reactions for the frame shown.

5

4

3

20 kN

25 kN

C m 4 10 kN B m 4 A

D

E 12 m

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Solution:

5 20 kN

20 kN

4

3 25 kN

15 kN

C m 4 10 kN B m 4 A HA

D

E

VA

VE

12 m

Resolve the 25 kN inclined external load into horizontal and vertical components as shown.

∑X =0, HA +10 =15 kN ⇒HA =5 kN Take moment about A, 10*4 + 20*12 – 15*8 – V E*12 =0 ⇒V E =13.3 kN

∑ Y =0, 20 +20 =V A +V E ⇒V A =26.7 kN

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Example 5 Find the reactions for the cantilever beam shown. 6 kN

4 kN/m 4 kNm

A

B

1.5m 1.5m 1.5m 1.5m

Solution: 6 kN

4 kN/m 4 kNm

MA

HA

VA

∑X =0,

A

B

1.5m 1.5m 1.5m 1.5m

HA =0 kN

∑ Y =0, 6 +4*1.5/2 =V A ⇒V A =9 kN Take moment about A, 6*1.5 + 4*(1.5/2)*(6 - 1.5*1/3) – 4 – M A =0 ⇒MA =21.5 kNm

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Example 6 Determine the support reactions for the frame shown.

8 kN/m 3 kN/m B

C

m / N k 3

m 2 1

A 8m

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Solution:

5 kN/m 3 kN/m

3 kN/m

B

C

m / N k 3

HA

m 2 1

A MA VA

∑X =0,

8m

HA =3*12 kN =36 kN

∑ Y =0, 3*8 +5*8/2 =V A ⇒V A =44 kN Take moment about A, 3*12*6 + 3*8*4 + 5*(8/2)*(8*2/3) – M A =0 ⇒MA =418.7 kNm

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Problems Determine the support reactions for the following structures. Q1.

8 KN/m A

B 3m

C 2m

Q2. 40 KN 12 KN/m

A

B 2m

C 2m

D 5m

Q3. 4 MN 2 MN/m

A

B 2m

C 2m

D 5m

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Q4.

500 KNm

40 KN/m A

B 6m

C 2m

D 2m

Q5. 500 Kg 800 kg/m

A

B C 500 mm

D 200 mm

100mm

Q6. 800 N 500 N 200 Nm A

B

C 500 mm

D 300 mm

200mm

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Q7.

250 N/m A

B 700 mm

C 500 mm

Q8. 1200 N 1500N/m

A

B 200mm

C 600 mm

D 500 mm

Q9. 800 N 600 N/m

A

B 700 mm

C

D

200 mm 300 mm

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Q10.

60 KN

150 KNm

A

B

C

4m

3m

D 2m

Q11. 800 kg 300 kg/m

A

B

C

2m

3m

Q12.

500 Nm

1500 N/m

A

B 1200m

C 700 mm

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Principle of Statics.pdf

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