Shear Force and Bending Moment Diagrams for a Simply Supported Beam

Every beam design starts with two numbers: the maximum shear force and the maximum bending moment. They decide the depth of the beam, the main reinforcement and the stirrups. Shear force and bending moment diagrams (SFD and BMD) show how these quantities vary along the span. This article explains the sign convention, the standard formulas for common load cases, and then solves a complete example of a simply supported beam carrying both a uniformly distributed load and a point load.

Definitions and Sign Convention

Shear force at a section is the algebraic sum of all vertical forces on one side of that section. Bending moment at a section is the algebraic sum of moments of all forces on one side of the section about that section. We will always take the left-hand side.

Shear force is positive when the resultant of forces to the left of the section acts upward.
Bending moment is positive (sagging) when it causes the beam to bend concave upward, with tension at the bottom.
Simply supported beams under downward loads have sagging moments throughout, so reinforcement goes at the bottom.

Relationships to remember

dV/dx = -w      (slope of SFD = intensity of load)
dM/dx =  V      (slope of BMD = shear force)

Consequences:
- Between point loads with no UDL, SF is constant and BM is a straight line.
- Under a UDL, SF varies linearly and BM is a parabola.
- Bending moment is maximum where shear force is zero or changes sign.

Standard Load Cases

Loading on span L	Reactions	Max shear	Max bending moment
Central point load P	P/2 each	P/2	PL/4 at centre
Point load P at distance a from A (b from B)	RA = Pb/L, RB = Pa/L	Larger reaction	Pab/L under the load
UDL w over full span	wL/2 each	wL/2	wL²/8 at centre

These results are worth memorising because the principle of superposition lets you add them for combined loads. However, for a combined load the maximum moment may not lie where any single case predicts, so it is safer to draw the full diagrams.

Worked Example

A simply supported beam AB has a span of 6 m. It carries a UDL of 10 kN/m over the full span and a point load of 30 kN at C, 2 m from A.

            30 kN
              |
   10 kN/m    v
  vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
  =====================================
  ^           C                       ^
  A <-- 2 m -->|<-------- 4 m -------> B
  RA                                  RB

Step 1: Support reactions

Total UDL = 10 x 6 = 60 kN acting at 3 m from A

Take moments about B (clockwise positive):
RA x 6 = 60 x 3 + 30 x 4 = 180 + 120 = 300
RA = 50 kN

ΣV = 0:  RA + RB = 60 + 30 = 90
RB = 90 - 50 = 40 kN

Step 2: Shear force

Section between A and C (0 < x < 2):
  V = 50 - 10x
  V(0) = 50 kN,  V(2-) = 50 - 20 = 30 kN

At C the 30 kN point load causes a sudden drop:
  V(2+) = 30 - 30 = 0 kN

Section between C and B (2 < x < 6):
  V = 50 - 10x - 30 = 20 - 10x
  V(6) = 20 - 60 = -40 kN  (equals -RB, check OK)

Shear force becomes zero exactly at C, at x = 2 m, so maximum bending moment occurs there.

Step 3: Bending moment

0 <= x <= 2:  M = 50x - 10x²/2 = 50x - 5x²
2 <= x <= 6:  M = 50x - 5x² - 30(x - 2)

M at C (x = 2) = 50 x 2 - 5 x 4 = 100 - 20 = 80 kNm
Check from the right: M = 40 x 4 - 10 x 4²/2 = 160 - 80 = 80 kNm  OK

x from A (m)	Shear force (kN)	Bending moment (kNm)
0	+50	0
1	+40	45
2 (left of C)	+30	80
2 (right of C)	0	80
3	-10	75
4	-20	60
5	-30	35
6	-40	0

Step 4: Sketch the diagrams

SFD (kN)
 +50 |\
     |  \ +30
     |   |
   0 |---+------------------------
     |    \
     |       \
     |          \
 -40 |             \_____________|
     A    C                      B

BMD (kNm), sagging drawn upward
  80 |      /‾‾‾\__
     |    /         ‾‾\__
     |  /                ‾‾\
   0 |/______________________\
     A    C                   B

The SFD is a sloping line from +50 to +30, a vertical drop of 30 at C, then another sloping line from 0 down to -40. The BMD is a curve rising to 80 kNm at C and falling back to zero at B. Notice that the BMD has a kink at C because the shear force jumps there.

Using the Results in Design

These are characteristic or working values. For limit state design per IS 456, multiply by the partial safety factor, typically 1.5 for dead plus live load. The design moment becomes Mu = 1.5 x 80 = 120 kNm and the design shear at support A becomes Vu = 1.5 x 50 = 75 kN. These values are then used to size the beam, calculate bottom steel and design shear reinforcement.

Common Mistakes

Taking the UDL resultant at the wrong point; it acts at the centre of the loaded length.
Forgetting the sudden jump in the SFD under a point load.
Assuming maximum moment is at mid-span for unsymmetrical loading.
Mixing sign conventions between the left and right sides of a section.

Summary

Find reactions using equilibrium, write shear and moment expressions for each segment, and plot them. Shear is the slope of the moment diagram, so maximum moment occurs where shear crosses zero. For the 6 m beam with 10 kN/m and a 30 kN load at 2 m, the reactions are 50 kN and 40 kN and the maximum moment is 80 kNm under the point load.