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

Page 1-1

PRINCIPLES OF STATICS Statics is the study of how forces act and react on rigid bodies which are at rest or not in motion. This study is the basis for the engineering principles, 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: For every action there is always an equal and opposite reaction. The study of statics has many applications in daily life. Anyone who has used a ladder should have a deep appreciation for the laws that govern the stability of the ladder. Hopefully the reactions generated by the ground surface and the wall surface are sufficient to maintain equilibrium. If these forces cannot be developed, an accident will likely occur.

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

Page 1-2

EQUILIBRIUM OF STRUCTURES A structure is considered to be in equilibrium if it remains at rest when subject to a system of forces and moments. If a structure is in equilibrium, then all its members and parts are also in equilibrium. For a structure to be in equilibrium, all the forces and moments (including support reactions) acting on it must balance each other. For a plane structure subject to forces in its own plane, the conditions for equilibrium can be expressed by the following equations of equilibrium: The third equation above states that the sum of moments of all forces about any point in the plane of the structure is zero.

∑ = ,0xF F1 + (-F2) = 0 F1 = F2 ∑ = ,0yF F3 + (-F4) = 0 F3 = F4

Forces and their resultants Forces are vectors, and because of this they have both magnitude and direction. This is extremely important to realize because an analysis of a structure cannot be completed by simply knowing the magnitude of forces without the direction of application.

∑ ∑ ∑ === 0 ,0 ,0 zyx MFF

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

Page 1-3

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.

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

Page 1-4

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

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

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

Page 1-5

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

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

Page 1-6

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. ∑ ∑ ∑ === 0 ,0 ,0 zyx MFF

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 equilibrium and prevent it from moving away under the loads, we must tie it up with some ties to the foundation.

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

Page 1-7

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 rotate about 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 rotate under the action of external loads, and it can be maintained in the conditions of equilibrium.

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

Page 1-8

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

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

Page 1-9

TYPES OF STRUCUTRES

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

Page 1-10

SUPPORTS A pin or hinge support is represented by the symbol

HB VBHB VB

or

A roller support is represented by the symbol

VB VB

or

Pin or Hinge Support

Prevented: Horizontal translation and vertical translation Allowed: Rotation

Roller Support

Prevented: Vertical translation Allowed: Horizontal translation and Rotation

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

Page 1-11

A fixed support is represented by the symbol

HB

VB

MBHB V

B MB

or

Fixed Support

Prevented: Horizontal translation, Vertical translation and Rotation Allowed: None

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

Page 1-12

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 kN20 kN

4 kN/mA

B C D

3m 3m1.5m 1.5m

Solution:

HA VA VC

20 kN20 kN

4 kN/mA

B C D

3m 3m1.5m 1.5m

∑X = 0, HA = 0 Take moment about A, 20*3 + 20*9 + 4*(4.5)*(4.5 + 4.5/2) – Vc*6 = 0 ⇒Vc = 60.25 kN ∑Y = 0, 20 + 20 + 4*4.5 = VA + VC

⇒VA = -2.25 kN, (-ve sign indicates VA acts in opposite direction) ⇒VA = 2.25 kN (↓)

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

Page 1-13

Example 2 Find the support reactions for the simple beam shown.

AB C

D

5m 2.5m 2.5m

40 kN 50 kN 3

45

Solution:

HA VA VD

AB C

D

5m 2.5m 2.5m

40 kN 50 kN3

45

40 kN

30 kN

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 – VD*10 = 0 ⇒VD = 50 kN ∑Y = 0, 40 + 40 = VA + VD

⇒VA = 30 kN

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

Page 1-14

Example 3 Determine the truss reaction forces.

5m 5m 5m 5m

5m

30 kN30 kN

20 kN

A B

Solution:

HA VA VB5m 5m 5m 5m

5m

30 kN30 kN

20 kN

A B

∑X = 0, HA = 20 kN Take moment about A, 30*10 + 30*15 – 20*2.5 – VB*20 = 0 ⇒VB = 35 kN ∑Y = 0, 30 + 30 = VA + VB

⇒VA = 25 kN

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

Page 1-15

Example 4 Determine the support reactions for the frame shown.

12 m

4m4m

20 kN

10 kN

25 kN3

4 5

A

B

C D

E

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

Page 1-16

Solution:

HA VA VE12 m

4m4m

20 kN

10 kN

25 kN3

4 5

A

B

C D

E

20 kN

15 kN

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 – VE*12 = 0 ⇒VE = 13.3 kN ∑Y = 0, 20 + 20 = VA + VE

⇒VA = 26.7 kN

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

Page 1-17

Example 5 Find the reactions for the cantilever beam shown.

1.5m 1.5m 1.5m 1.5m

6 kN4 kNm

4 kN/m

A B

Solution:

HA

VA

MA

1.5m 1.5m 1.5m 1.5m

6 kN4 kNm

4 kN/m

A B

∑X = 0, HA = 0 kN

∑Y = 0, 6 + 4*1.5/2 = VA ⇒VA = 9 kN

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

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

Page 1-18

Example 6 Determine the support reactions for the frame shown.

8m

12m

3 kN/m3

kN/m

8 kN/m

A

B C

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

Page 1-19

Solution:

HAVA

MA

8m

12m

3 kN/m

3 kN

/m

5 kN/m

A

B C

3 kN/m

∑X = 0, HA = 3*12 kN = 36 kN

∑Y = 0, 3*8 + 5*8/2 = VA ⇒VA = 44 kN

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

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

Page 1-20

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

Q2.

Q3.

A B C

2 MN/m

2 m 5 m

4 MN

D

2 m

A B C

12 KN/m

2 m 5 m

40 KN

D

2 m

A B C

8 KN/m

3 m 2 m

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

Page 1-21

Q4.

Q5.

Q6.

B C

800 N

500 mm 200mm

200 Nm

D 300 mm

500 N

A

B C

500 Kg

500 mm 100mm

800 kg/m

D 200 mm

A

A B C

500 KNm

6 m 2 m

40 KN/m

D 2 m

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

Page 1-22

Q7.

Q8.

Q9.

A B C

800 N

700 mm 200 mm

600 N/m

D

300 mm

A B C

1500N/m

200mm 500 mm

1200 N

D

600 mm

A B C

250 N/m

700 mm 500 mm

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

Page 1-23

Q10.

Q11.

Q12.

B 700 mm 1200m

500 Nm

C A

1500 N/m

B

800 kg

2 m 3 m

300 kg/m

C A

A B C

60 KN

4 m 3 m

150 KNm

D 2 m