Mechanics of Materials

1. Mechanics of Materials
CE 1201
Md. Hossain Nadim
Dept. of Civil Engineering

2. Computation of Reactions
Md. Hossain Nadim, Lecturer, Dept. of CE, AUST

3. Idealized Structure: An exact analysis of a structure can never be carried out, since estimates always
have to be made of the loadings and the strength of the materials composing the structure.
Furthermore, points of application for the loadings must also be estimated. It is important,
therefore, that the structural engineer develop the ability to model or idealize a structure so that he
or she can perform a practical force analysis of the members.

4. Support Connections: Structural members are joined together in various ways depending on the
intent of the designer. The three types of joints most often specified are the pin connection, the
roller support, and the fixed joint. A pin-connected joint and a roller support allow some freedom
for slight rotation, whereas a fixed joint allows no relative rotation between the connected
members.
Pin/Hinge Support

5. Fixed Support

6. Roller Support

7. Idealized models used in structural analysis that represent pinned and fixed supports and pin-
connected and fixed-connected joints are shown in the following figures:

8. Supports for coplanar structures:
Roller Suppport One unknown.
The reaction is a force
that acts perpendicular to
the surface at the point of
contact.
Hinge Support Two unknowns.
The reactions are two
force components.
Fixed Support Three unknowns.
The reactions are the
moment and the two
force components.

9. Idealization of Loading:
P
Actual Idealized
Actual Idealized
Actual Idealized

10. Types of Loading:
1. Point Load
2. Distributed Load
• Uniformly distributed load: Uniformly distributed load is that whose magnitude remains uniform
throughout the length.
• Uniformly varying load: It is that load whose magnitude varies along the loading length with a
constant rate.
P

11. Uniformly distributed load
(UDL)
L
Uniformly Varying load
(UVL)

12. 3. Combined

13. Equation of Equilibrium:
It may be recalled from statics that a structure or one of its members is in equilibrium
when it maintains a balance of force and moment. In general, this requires that the force
and moment equations of equilibrium be satisfied along three independent axes, namely,
𝛴𝐹𝑥 = 0; 𝛴𝐹𝑦 = 0;𝛴𝐹𝑧 = 0
𝛴𝑀 𝜒 = 0; 𝛴𝑀 𝑦 = 0; 𝛴𝑀𝑧 = 0
The principal load-carrying portions of most structures, however, lie in a single plane,
and since the loads are also coplanar, the above requirements for equilibrium reduce to
𝛴𝐹𝑥 = 0
𝛴𝐹𝑦 = 0
𝛴𝑀 = 0

15. Types of Structure
Stability and Determinacy:
 Determinate Structure: A structure for which all the unknown reactions can be determined using
the equations of equilibrium is referred to as a determinate structure.
 Indeterminate Structure: The structure possessing more unknown reactions than equations of
equilibrium, are referred to as indeterminate structure.
𝑟 = 3𝑛, statically determinate
𝑟 > 3𝑛, statically indeterminate

16. Unstable Structure: a structure will be geometrically unstable—that is, it will move slightly or
collapse—if there are fewer reactive forces than equations of equilibrium; or if there are enough
reactions, instability will occur if the lines of action of the reactive forces intersect at a common point
or are parallel to one another.
L/2 L/2
P
Partially restrained
Concurrent forces
Parallel forces

17. Steps for determining Reactions:
1. Assuming Reactions with direction
2. Apply σ 𝑴 𝒂𝒏𝒚 𝒑𝒐𝒊𝒏𝒕 = 𝟎
3. Apply σ 𝑭 𝒚 = 𝟎
4. Apply σ 𝑭 𝒙 = 𝟎 (if axial force exist)

18. Example:1
Determine the reactions for a simply supported beam subjected to three
point loads as shown in the Fig. given below.

19. 𝑅 𝐴𝑥
𝑅 𝐴𝑦
𝑅 𝐵𝑦
Using the condition: + 𝛴𝑀𝐴 = 0;
– RBy × 8 + 8 × 7 + 10 × 4 + 5 × 2 = 0  RBy = 13.25 N
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 5 – 10 – 8 = 0  RAy = 9.75 N
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  RAx = 0 N

20. 5 ft 5 ft 5 ft
5 kip/ft
10 kips
5
12
Example: 2
Determine the reactions for a simply supported beam subjected to the
loads as shown in the Fig. given below.

21. P P
P P
P
θ
P
1
1
L
w
wL
L/2 L/2

22. 5 ft 5 ft 5 ft
5 kip/ft
10 kips
5
12
Solution:
𝑅 𝐴𝑥
𝑅 𝐴𝑦 𝑅 𝐵𝑦
13
0
5
3
= 3.85 𝑘𝑖𝑝
0
3
= 9. 3 𝑘𝑖𝑝
5 5 = 5 𝑘𝑖𝑝
Using the condition: + 𝛴𝑀𝐴 = 0;
3.85 × 5 + 25 × 12.5 − RBy × 15 = 0  RBy = 22.12 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 3.85 – 25 = 0;  RAy = 6.73 kip
Using the condition: ՜
+
𝛴𝐹𝑥 = 0
RAx – 9.223 = 0  RAx = 9.23 Kip
2.5 ft 2.5 ft
A B

23. Example: 3
Determine the reactions for the double sided overhanging beam
subjected to the loads as shown in the Fig. given below.
3 kip/ft
6 kip/ft 10 kip-ft
3 ft 3 ft 6 ft5 ft

24. L
w
𝑤 𝐿
3
𝐿
3
𝐿

25. 3 kip/ft
6 kip/ft 10 kip-ft
3 ft 3 ft 6 ft5 ft
Solution:
𝑅 𝐴𝑥
𝑅 𝐴𝑦 𝑅 𝐵𝑦
𝟔 𝟓 = 𝟑𝟎 𝒌𝒊𝒑
𝟏
𝟐
𝟑 𝟔 = 𝟗 𝒌𝒊𝒑
2.5 ft2.5 ft 𝟐
𝟑
𝟔 = 𝟒
𝟏
𝟑
𝟔 = 𝟐
Using the condition: + 𝛴𝑀𝐴 = 0;
− 30 × 2.5 + 10 − RBy × 6 + 9 ×10 = 0  RBy = 4.17 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 30 – 9 = 0;  RAy = 34.83 kip
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  RAx = 0 Kip
A B
N.B: Don’t multiply distance with moment
N.B: Don’t insert moment in force equation

26. Example: 4
Determine the reactions for a fixed supported beam subjected to the
loads as shown in the Fig. given below.
5 kip
2 kip/ft
B
20 kip10 kip
A
3 ft 2 ft1 ft

27. Using the condition: +↑ 𝛴𝐹𝑦 = 0;
− 5 −10 −20 −12 + RBy = 0;  RBy = 47 kip
Using the condition: + 𝛴𝑀 𝐵 = 0;
−5 × 6 −20 × 2 − 22 × 3− 𝑀 = 0  M = −136 kip-ft =136 kip-ft
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  RAx = 0 Kip
Solution:
5 kip 2 kip/ft
B
20 kip10 kip
A 3 ft 2 ft1 ft
𝑅 𝐵𝑥
𝑅 𝐵𝑦
𝑀
𝟐 𝟔 = 𝟏𝟐 𝒌𝒊𝒑
3 ft3 ft
N.B: Moment (M) has to be included

28. Exercise problems:
1.
2.
8 ft 3 ft
20 Kip
40 kip/ft
150 kip-ft

29. Example: 4
Determine the reactions for the single side overhanging beam
subjected to loading as shown in the Fig. given below
30 kip/ft
20 kip/ft
1 ft 1 ft 2 ft2 ft
40 kip
0.7 ft
0.5 ft

30. Solution:
30 kip/ft
20 kip/ft
1 ft 1 ft 2 ft2 ft
40 kip
40 x 0.5 =
20 kip-ft
1 ft1 ft
BA
𝑅 𝐴𝑥
𝑅 𝐴𝑦 𝑅 𝐵𝑦
𝟐𝟎 𝟐 = 𝟒𝟎 𝒌𝒊𝒑
𝟏
𝟐
𝟐 𝟑𝟎 = 𝟑𝟎 𝒌𝒊𝒑
𝟏
𝟑
𝟐 =
𝟐
𝟑
𝟐
𝟑
𝟐 =
𝟒
𝟑Using the condition: + 𝛴𝑀𝐴 = 0;
40 + 40 3 + 20 − RBy × 4 + 30 ×(4+
2
3
) = 0  RBy = 80 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 40 – 40−30 = 0;  RAy = 30 kip
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  RAx = 0 Kip

31. Exercise problems:
20kN/m
40kN
3m 1m
A
1m
0.7m
0.5m
300
DB C

32. Special Case: Internal Hinge
Internal
Hinge
2 m2 m
2 m

34. Example 5:
Determine the reactions for the compound beam with internal
hinge at C point, is subjected to loading as shown in the Fig.
given below.

35. Solution:
CC
𝐶 𝑥
𝐶 𝑦
𝐶 𝑥
𝐶 𝑦

36. C
𝑅 𝐴𝑥
𝑅 𝐴𝑦
Using the condition: + 𝛴𝑀 𝐶 = 0;
4 0 0.5 − RDy × 9  RDy = 490 kN
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
Cy + RDy – 420 = 0;  Cy = −70 kip = 70 kN
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  Cx = 0
C 𝐶 𝑥
𝐶 𝑦
𝑅 𝐷𝑦
10.5 m10.5 m
𝟐𝟎 𝟐𝟏 = 𝟒𝟐𝟎 𝒌𝑵
𝑀
𝐶 𝑥 = 0
𝐶 𝑦 = 70 k p
Figure-1
Figure-2
N.B From Fig.1 𝐶 𝑦 actually works downward so in Fig.2 𝐶 𝑦 has to work upwards
Using the condition: + 𝛴𝑀𝐴 = 0;
−𝑀 + 30 − 70 4 = 0;  M = 1080 kN-m
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
𝑅 𝐴𝑦 − 30 + 70 = 0 ;  RAy = 160 kN-m
Using the condition: ՜
+
𝛴𝐹𝑥 = 0  𝑅 𝐴𝑥 − Cx = 0; 𝑅 𝐴𝑥 =0

37. Ans:
Exercise:
1 kip/ft 1 kip/ft
6 Kip
4 ft 3 ft 10 ft10 ft 3 ft 4 ft
Hinge

38. More Exercise Problems:
200N 100N
A B
30N/m
4m 4m3m