Mechanics of Materials

1. Mechanics of Materials
CE 1201
Md. Hossain Nadim
Dept. of Civil Engineering
Lecture-2

2. Shear Force and Bending Moment
Md. Hossain Nadim, Lecturer, Dept. of CE, AUST

3. Internal Loadings Developed in Structural Members:
If the internal loadings acting on the cross section at point B are to be determined, we must pass an imaginary
section a–a perpendicular to the axis of the beam through point B and then separate the beam into two
segments. The internal loadings acting at B will then be exposed
The force component 𝑁 𝐵 that acts perpendicular to the cross section is termed the normal force (Axial force).
The force component 𝑉𝐵 that is tangent to the cross section is called the shear force, and the couple moment
𝑀 𝐵 is referred to as the bending moment.

4. Shear force: Parallel to the cross-section
Axial force: Perpendicular to the cross-section
Shear Force and Bending Moment Diagram
Because of the applied loadings, beams develop an internal shear force and bending moment that, in general, vary from point
to point along the axis of the beam. Shear and moment diagrams are plots of internal shear force and internal bending moment
as a function of x. By looking at these plots, we can immediately see the maximum values of the shear force and the bending
moment, as well as the location of these maximum values.

5. Sign Convention
Positive axial force
Tension
Negative axial force
Compression
Negative ShearPositive Shear
Fiber in
Compression
Fiber in Tension Fiber in
Compression
Fiber in Tension

6. Relationships Between Load, Shear, And Bending Moment:
Let us consider the beam shown in Fig. (a), which is subjected to an arbitrary loading. A free‐body diagram for a small
segment Δx of the beam is shown in Fig.(b). All the loadings shown on the segment act in their positive directions
according to the established sign convention. Also, both the internal resultant shear and moment, acting on the right face of
the segment, must be changed by a small amount in order to keep the segment in equilibrium. The distributed load has
been replaced by a resultant force 𝑤(𝑥)Δ𝑥 that acts at a fractional distance 𝑘(𝛥𝑥) from the right side, where 0 < 𝑘 < 1
[for example, if 𝑤(𝑥) is uniform, k ꞊ 0.5].
F
Mo
x Δx
w(x)
(a)
M M + ΔM
V
V +ΔV
Δx
w(x)
k(Δ x)
w(x)Δ x
o
(b)

7. Applying the equations of equilibrium to the segment, we have
+↑ 𝛴𝐹𝑦 = 0; ⇒ 𝑉 + 𝑤(𝑥)Δ𝑥 − (𝑉 + Δ𝑉) = 0
⇒ 𝛥𝑉 = 𝑤(𝑥)Δ𝑥
⤹ +𝛴𝑀 𝑂 = 0; ⇒ −V Δ𝑥 − 𝑀 − 𝑤(𝑥)Δ𝑥[𝑘(𝛥𝑥)] + (𝑀 + Δ𝑀) = 0
⇒ 𝛥𝑀 = 𝑉Δ𝑥 + 𝑤(𝑥)𝑘(Δ𝑥)2
Dividing by Δ𝑥 and taking the limit as Δ𝑥 → 0, the above two equations become
𝑑𝑉
𝑑𝑥
꞊ 𝑤(𝑥)
∴ slope of distributed shear diagram at each point = load intensity at each point
𝑑𝑀
𝑑𝑥
꞊ 𝑉
∴ slope of moment diagram at each point = shear at each point
These two equations provide a convenient means for quickly obtaining the shear and moment
diagram for a beam

8. Using the relationships:
𝟐𝟎′
P = 20 kips
𝟏𝟐′
SFD
(kips)
8 kips 12 kips
8
12
+
-
BMD
(kip-ft)
96
+
Load, w = 0
Shear, V = ‫׬‬ 𝑤 𝑑𝑥
= ‫׬‬ 0 𝑑𝑥
= C
Moment = ‫׬‬ 𝑉 𝑑𝑥
= ‫׬‬ 𝐶 𝑑𝑥
= C x +𝐶1
It is clear that order of x increase
with every step

9. Using these equations the following common relationship can be developed

11. +i
+c
+d
-i
-c
-d
i = Increase
c = Constant
d = Decrease
Important Notes:
• The slope of the shear diagram at a point is equal to the intensity of the
distributed loading, where positive distributed loading is upward, i. e.,
𝑑𝑉/𝑑𝑥 = 𝑤(𝑥) .
• The change in the shear 𝛥𝑉 between two points is equal to the area under
the distributed‐loading curve between the points.
• If a concentrated force acts upward on the beam, the shear will jump
upward by the same amount.
• The slope of the moment diagram at a point is equal to the shear, i.e.,
𝑑𝑀/𝑑𝑥 = 𝑉.
• The change in the moment 𝛥𝑀 between two points is equal to the area
under the shear diagram between the two points.
• If a clockwise couple moment acts on the beam, the shear will not be
affected; however, the moment diagram will jump upward by the amount of
the moment.
• Points of zero shear represent points of maximum or minimum moment
since 𝑑𝑀/𝑑𝑥 = 0.
• Because two integrations of 𝑤 = 𝑤(𝑥) are involved to first determine the
change in shear, 𝛥𝑉 = ‫׬‬ 𝑤 (𝑥)𝑑𝑥, then to determine the change in moment,
𝛥𝑀 = ‫׬‬ 𝑉 𝑑𝑥, then if the loading curve 𝑤 = 𝑤(𝑥) is a polynomial of
degree 𝑛, 𝑉 = 𝑉(𝑥) will be a curve of degree 𝑛 + 1, and 𝑀 = 𝑀(𝑥) will be
a curve of degree 𝑛 + 2.

12. 𝟐𝟎′
P = 20 kips
𝟏𝟐′
Example (1): Draw the SFD and BMD for the following beam subjected to indicated loadings.

13. 𝟐𝟎′
P = 20 kips
𝟏𝟐′
SFD
(kips)
8 kips 12 kips
8
12
+
-
BMD
(kip-ft)
𝟖 × 𝟏𝟐 = 𝟗𝟔
+
Using the condition: + 𝛴𝑀𝐴 = 0;
20 × 12 - RBy × 20 = 0
⇒ RBy = 12 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 20 = 0;
⇒ RAy = 8 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦

14. Example (2): Draw the AFD, SFD and BMD for the following beam
𝟐𝟎′
P = 20 kips
𝟏𝟐′
3
4

15. Using the condition: + 𝛴𝑀𝐴 = 0;
12× 12 - RBy × 20 = 0
⇒ RBy = 7.20 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 12 = 0;
⇒ RAy = 4.8 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 16
𝟐𝟎′
P = 20 kips
𝟏𝟐′
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦
3
4
5
12 kip
16 kip
SFD
(kips)
4.8
7.2
+
-
BMD
(kip-ft)
𝟒. 𝟖 × 𝟏𝟐 = 𝟓𝟕. 𝟔
+
-16
AFD
(kips)

16. 10 ft
2 kip/ft
Example (3): Draw the SFD and BMD for the following beam

17. 10 ft
2 kip/ft
A𝑅 𝐴𝑥
𝑅 𝐴𝑦 =10
B
𝑅 𝐵𝑦 = 10
Using the condition: + 𝛴𝑀𝐴 = 0;
20 × 5 + RBy × 10 = 0
⇒ RBy = 10 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + RBy – 20 = 0;
⇒ RAy = 10 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
10 × 2 = 20 𝑘𝑖𝑝
5 ft 5 ft
10
10
+
-
0.5 × 5 × 10 = 𝟐𝟓
SFD
(kips)
BMD
(kip-ft)
+ +

18. Example (4): Draw the SFD and BMD for the following beam. Neglect self weight of the beam
2 ft 2 ft
10 Kip 10 Kip

19. 2 ft 2 ft
10 Kip 10 Kip
𝑅 𝐴𝑥
𝑅 𝐴𝑦
𝑀
Using the condition: + 𝛴𝑀𝐴 = 0;
−𝑀 +10 × 2 - 10 × 4 = 0
⇒ M = 60 kip-ft
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy – 10 – 10 = 0;
⇒ RAy = 20 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
20
+
10
+
60
20
0
= 20 − 10 × 2
= 60 − 20 × 2
SFD
(kips)
BMD
(kip-ft)
-
-

20. Example (4): Draw the SFD and BMD for the following beam. Neglect self weight of the beam
1.5 ft
0.5 kip/ft
2 Kip 4 Kip
1.5 ft 1 ft

21. 4 × 0.5 = 2 𝑘𝑖𝑝
1.5 ft
0.5 kip/ft
2 Kip 4 Kip
1.5 ft 1 ft
A𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦
Using the condition: + 𝛴𝑀𝐴 = 0;
2×1.5 + 4×3+2×2-𝑅 𝐵𝑦× 4= 0
⇒ 𝑅 𝐵𝑦 = 4.75 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + 𝑅 𝐵𝑦 – 2 – 4 –2 = 0;
⇒ RAy = 3.25 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
2 ft 2 ft

22. 1.5 ft
0.5 kip/ft
2 Kip 4 Kip
1.5 ft 1 ft
3.25
+
+
SFD
(kips)
BMD
(kip-ft)
-
A
3.25 K
B
4.75 K
-
2.5=(3.25−0.5x1.5)
0.5
0.25
4.25
4.75
4.31
4.56
4.5
x
1.5-x Taking the similar triangles:
𝑥
0.5
=
1.5 − 𝑥
0.25
⇒ 𝑥 = 1
0.5x(3.25+2.5)x1.5=

23. Example (5): Draw the SFD and BMD for the following beam. Neglect self weight of the beam
90 kip/ft
20 kip/ft
2 ft 2 ft 1 ft2 ft
200 Kip
3
4
2 ft 1 ft

24. 90 kip/ft
20 kip/ft
2 ft 2 ft 1 ft2 ft
200 Kip
3
4
2 ft 1 ft
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦
Using the condition: + 𝛴𝑀𝐴 = 0;
-90× (
1
3
× 2) + 160×2-𝑅 𝐵𝑦× 4+40 × 6= 0
⇒ 𝑅 𝐵𝑦 = 125 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + 𝑅 𝐵𝑦 – 90 – 160 –40 = 0;
⇒ RAy = 165 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 120 kip
160 K
120 K
0.5x90x2=90 Kip
20x2=40 Kip
1
3
x 2
2
3
x 2
11

25. 90 kip/ft
20 kip/ft
2 ft 2 ft 1 ft2 ft
200 Kip
3
4
2 ft 1 ft
A
165 K
B
125 K
120
120 -
-
-
+ + +
90
75
85
40
60
90
80 40
- -
- -
+ +
SFD
(kips)
BMD
(kip-ft)
AFD
(kips) 𝑏
𝑎
2
3
𝑎𝑏
𝑎
𝑏
1
3
𝑎𝑏
160 K
120 K

26. 2 ft
2 kip/ft
10 kip-ft
2 ft
Example (6): Draw the SFD and BMD for the following beam. Neglect self weight of the beam

27. 2 ft
2 kip/ft
10 kip-ft
2 ft
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦
Using the condition: + 𝛴𝑀𝐴 = 0;
8× 2 + 10-𝑅 𝐵𝑦× 4= 0
⇒ 𝑅 𝐵𝑦 = 6.5 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + 𝑅 𝐵𝑦 – 8 = 0;
⇒ RAy = 1.5 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
4x2=8 Kip

28. 2 ft
2 kip/ft
10 kip-ft
2 ft
SFD
(kips)
BMD
(kip-ft)
A
1.5 K
B
6.5 K
1.5
6.5
-
-
+
𝑥 = 0.75
4 − 𝑥 =3.25
1.5
𝑥
=
6.5
4 − 𝑥
⇒ 𝑥 = 0.75
1.5
0.75
=
𝑦
1.25
⇒ 𝑦 = 2.5
𝑦
1.25 2
0.56
1
9

29. 4 ft
3 kip/ft
Example (7): Draw the SFD and BMD for the following beam. Neglect self weight of the beam

30. 4 ft
3 kip/ft
Using the condition: + 𝛴𝑀𝐴 = 0;
6 x
8
3
-𝑅 𝐵𝑦× 4= 0
⇒ 𝑅 𝐵𝑦 = 4 kip
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
RAy + 𝑅 𝐵𝑦 – 6 = 0;
⇒ RAy = 2 kip
Using the condition: →
+
𝛴𝐹𝑥 = 0
RAx = 0
0.5x4x3=6 Kip
2
3
×4 ft
1
3
×4 ft
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
B
𝑅 𝐵𝑦

31. 4 ft
3 kip/ft
4 ft
3 kip/ft
A
2 K
B
6.5 K
SFD
(kips)
2
4
𝑥
A
2 K
B
6.5 K
3
4
𝑥
𝑥2 K 𝑉
𝑀
Using the condition: +↑ 𝛴𝐹𝑦 = 0;
⇒ 2 + 𝑉–
3
8
𝑥2
= 0;
⇒ V =
3
8
𝑥2
-2
When, V=0; ⇒
3
8
𝑥2-2 =0 ⇒ 𝑥 = 2.31
0.5 (
3
4
𝑥)(𝑥 )
𝑥 = 2.31
3.08
BMD
(kip-ft)

32. More Practice Problems (with solution):

36. Calculate the value and draw a bending moment and shear
force diagram for following beam shown in fig also find the
point of contra-flexure.

38. Exercise Problems:
1. The cantilever beam AC in Fig. loaded by the uniform load of 600 Kip/ft over
the length BC together with the moment of magnitude 4800 kip-ft working at
the tip C. Determine the shearing force and bending moment diagrams.
2 ft 2 ft
600 kip/ft
4800 kip-ft
A
B
C

39. 2. The beam AC is simply supported at A and C and subject to the uniformly
distributed load of 300 Kip/ft from B to C plus a moment of magnitude 2700
kip-ft is applied at D as shown in Fig., draw the shearing force and bending
moment diagrams.
300 kip/ft
3 ft 3 ft3 ft
2700 kip-ft
A
B C
D

40. 3. The overhanging beam AE is subject to uniform normal loadings in the
regions AB and DE, together with a moment acting at the midpoint C as
shown in Fig., plot the shear and moment diagram.
4000 kip/ft
1.5 ft 1 ft1 ft
4000 kip/ft
200 kip-ft
1.5 ft
A B D EC