The document discusses shear force and bending moment diagrams. It defines shear force and bending moment, explaining that shear force acts perpendicular to the beam's axis while bending moment acts to bend the beam. It outlines the procedure to determine shear force and bending moment diagrams: (1) calculate support reactions, (2) divide the beam into segments based on loading, (3) draw free body diagrams and calculate expressions for each segment. As an example, it analyzes a simply supported beam with two loads to derive the shear force and bending moment expressions and diagrams.

1. Shear force and bending
moment diagrams
By Muhammad Bilal
CED, UET Peshawar

2. Introduction
• Shear force is the force in the beam acting perpendicular to its
longitudinal (x) axis. For design purposes, the beam's ability to
resist shear force is more important than its ability to resist an
axial force. Axial force is the force in the beam acting parallel to the
longitudinal axis
• Bending moment is just another kind of force that you apply on a
member.
• If you try to elongate a member by pulling its ends away form each
other, the force is tension.
• If you try to compress a member as if you want to reduce its length,
then the force is compression.
• If you try to Slice the member as if you were slicing bread, then
the force is Shear.
• Like wise, if you try to "Bend" a member, or rotate it then the force
is Bending moment.

3. Introduction(Cont…)
• In a beam, the internal force system consist of a shear force
and a bending moment acting on the cross section of the bar.
The shear force and the bending moment usually vary
continuously along the length of the beam
• The internal forces give rise to two kinds of stresses on a
transverse section of a beam: (1) normal stress that is caused
by bending moment and (2) shear stress due to the shear
force
• Knowing the distribution of the shear force and the bending
moment in a beam is essential for the computation of stresses
and deformations.

4. Procedure for determining shear force and
bending moment diagrams
• Compute the support reactions from the free-body
diagram (FBD) of the entire beam.
• Divide the beam into segment so that the loading
within each segment is continuous.
• Draw a FBD for the part of the beam lying either to
the left or to the right of the cutting plane,
whichever is more convenient. At the cut section,
show V and M acting in their positive directions.
• Determine the expressions for Shear force (V) and M
from the equilibrium equations obtain from the FBD

5. Cont…
• It is visually desirable to draw the shear force
V-diagram below the FBD of the entire beam, and
then draw the M- diagram below the V-diagram

6. Sample problem
The simply supported beam in Fig. (a) carries two concentrated
loads. (1) Derive the expressions for the shear force and the
bending moment for each segment of the beam. (2) Sketch the
shear force and bending moment diagrams. Neglect the weight
of the beam. Note that the support reactions at A and D have
been computed and are shown in Fig. (a).

7. Solution:
Part 1
The determination of the expressions for V and M for each of the
three beam segments (AB,BC, and CD) is explained below
Segment AB (0＜x＜2 m)
ΣFy =0 +↑
18-V = 0
V = +18 kN
ΣM(at E) = 0 (CCW +)
- 18x+ M = 0
M = +18x kN· m
The point E is at
Position 1 in fig(a)

8. Segment BC ( 2<x<5)
ΣFy =0 +↑
18-14-V = 0
V = +18-14 =
+4 kN Answer
ΣM(at f) = 0 CCW +
- 18x + 14(x-2) + M = 0
M = +18x-14(x-2)
= 4x+28 kN· m
Point F is at position 2 in fig a

9. Segment CD (5 m＜x＜7 m)
ΣFy =0 +↑
18-14—28-V = 0
V = +18-14-28
= -24 kN Answer
ΣM(at G) = 0 CCW +
-18x+ 14(x-2)+28(x-5)+M = 0
M = +18x-14(x-2) – (x-5)
= -24x+168 kN· m
Point G is at position 3 in fig a

10. Part 2
Shear force and bending moment diagram

11. Important points
• The V-diagram reveals that the largest shear
force in the beam is -24 kN : segment CD
• The M-diagram reveals that the maximum
bending moment is +48 kN·m : the 28-kN load
at C.