Interactive Lecture
Reactions, shear force, and bending moment in loaded beams
by Dr Arnaldo Delli Carri
Beams are structural elements that support loads primarily through bending. How a beam responds to different loading scenarios drives much of mechanical and civil engineering design.
This interactive lecture explores:
A simply supported beam rests on a pin at one end and a roller at the other. Together the supports:
These beams are statically determinate with two reaction forces.
A cantilever beam is fixed at one end and free at the other. The fixed support:
These beams are also statically determinate with one reaction force and one reaction moment.
For a beam to be in static equilibrium, the following conditions must be satisfied:
Sum of all forces in the vertical direction must equal zero:
\[\sum F_y = 0\]
Sum of all moments about any point must equal zero:
\[\sum M = 0\]
These principles allow us to determine the reaction forces at the supports when external loads are applied to the beam.
For a simply supported beam of length L with a concentrated force F at distance a from the left support:
Vertical reaction at right support (B):
\[B_y = -\frac{F \cdot a + M}{L}\]
Vertical reaction at left support (A):
\[A_y = -F - B_y\]
For a cantilever beam of length L with a concentrated force F at distance a from the fixed end:
Moment reaction at the fixed end (A):
\[M_a = -(F \cdot a + M)\]
Vertical reaction at the fixed end (A):
\[A_y = -F\]
We can determine the internal loads in beams by considering the effect of all forces and moments on either side of a section.
The shear force at any section is the sum of all transverse forces acting on either side of the section.
\[V(x) = \sum \text{forces to the left of section } x\]
The shear force diagram (SFD) shows how the shear force varies along the length of the beam.
The bending moment at any section is the sum of the moments of all forces acting on either side of the section.
\[M(x) = \sum \text{moments to the left of section } x\]
The bending moment diagram (BMD) shows how the bending moment varies along the length of the beam.
Positive N, V, and M are defined by their direction on a right-facing cut face. The diagram below shows the complete positive sign convention applied simultaneously at both exposed faces: the left and right beam segments show the resultants on their cut faces, and the isolated cross-section at the centre confirms that the two sets are equal and opposite — Newton's third law at the cut.
On a right-facing face: positive \(N\) acts to the right (tension in blue), positive \(V\) acts downward (green), and positive \(M\) acts clockwise (red) — producing sagging. On the left-facing face all three reverse.
For a simply supported beam, we can determine the shear force and bending moment by summing all forces and moments to the left of a given section.
For a cantilever beam, we can determine the shear force and bending moment by summing all forces and moments to the left of a given section, including the fixed-end reactions.
The derivative of the bending moment is equal to the shear force
\[\frac{dM(x)}{dx} = V(x)\]
The derivative of the shear force is equal to the negative of the distributed load
\[\frac{dV(x)}{dx} = -q(x)\]
Read together, a distributed load sets the slope of the shear diagram, and the shear sets the slope of the moment diagram.
Design Considerations: Key factors in beam design based on internal load analysis:
Internal load analysis runs through structural and mechanical engineering:
| Concept | Equation | Description |
|---|---|---|
| Vertical force equilibrium | \(\sum F_y = 0\) | \(F_y\) = vertical forces |
| Moment equilibrium | \(\sum M = 0\) | \(M\) = moments about any point |
| Shear force at section x | \(V(x) = \sum \text{forces to the left of } x\) | \(V(x)\) = shear force at position x |
| Bending moment at section x | \(M(x) = \sum \text{moments to the left of } x\) | \(M(x)\) = bending moment at position x |
| Derivative of bending moment | \(\frac{dM(x)}{dx} = V(x)\) | Shear force equals slope of moment diagram |
| Derivative of shear force | \(\frac{dV(x)}{dx} = -q(x)\) | \(q(x)\) = distributed load intensity |
A simply supported beam of length L = 6 m has a point load P = 12 kN applied 2 m from the left support. Calculate:
Show Solution
(a) Reaction forces:
Taking moments about the left support (A):
\[\sum M_A = 0: R_B \times 6 - 12 \times 2 = 0\]
\[R_B = \frac{24}{6} = 4 \text{ kN}\]
Vertical equilibrium:
\[\sum F_y = 0: R_A + R_B - 12 = 0\]
\[R_A = 12 - 4 = 8 \text{ kN}\]
(b) Shear force at point load:
Immediately to the left of P: \(V = R_A = 8 \text{ kN}\) (positive)
Immediately to the right of P: \(V = R_A - P = 8 - 12 = -4 \text{ kN}\) (negative)
The shear force has a step change of 12 kN at the load location.
(c) Maximum bending moment:
The maximum bending moment occurs where V = 0, which is at x = 2 m (at the load):
\[M_{max} = R_A \times 2 = 8 \times 2 = 16 \text{ kN·m}\]
Alternative calculation: \(M_{max} = R_B \times 4 = 4 \times 4 = 16 \text{ kN·m}\)
Answer: (a) R_A = 8 kN, R_B = 4 kN; (b) V = +8 kN to the left, −4 kN to the right of the load; (c) M_max = 16 kN·m at x = 2 m.
A cantilever beam of length L = 4 m is subjected to a uniformly distributed load w = 5 kN/m over its entire length. Determine:
Show Solution
(a) Reactions at fixed support:
Total load: \(W = w \times L = 5 \times 4 = 20 \text{ kN}\)
Vertical reaction: \(R_A = W = 20 \text{ kN}\) (upward)
Moment reaction: \(M_A = W \times \frac{L}{2} = 20 \times 2 = 40 \text{ kN·m}\) (counterclockwise)
(b) Shear force at x = 1 m from the fixed end:
The shear equals the net transverse force to one side of the cut. Measuring from the fixed end (forces to the left), with upward taken positive:
\[V = R_A - w \times 1 = 20 - 5 \times 1 = 15 \text{ kN}\]
Measuring from the free end (forces to the right) gives the same value — the remaining 3 m of downward load carries positive shear across the cut:
\[V = w \times (L - 1) = 5 \times 3 = 15 \text{ kN}\]
Both routes agree: \(V = +15 \text{ kN}\). As a check, at the free end \(V(L) = R_A - wL = 20 - 5 \times 4 = 0\), as required.
(c) Bending moment at fixed end:
\[M_{fixed} = -\frac{wL^2}{2} = -\frac{5 \times 4^2}{2} = -40 \text{ kN·m}\]
The negative sign indicates the moment creates tension on the top fibres (hogging moment).
Answer: (a) R_A = 20 kN upward, M_A = 40 kN·m; (b) V = +15 kN at 1 m from the fixed end; (c) M_fixed = −40 kN·m (hogging).
A simply supported beam of length L = 5 m is subjected to:
Calculate:
Show Solution
(a) Reaction forces:
Total UDL: \(W = 4 \times 5 = 20 \text{ kN}\) acting at centre (2.5 m)
By symmetry, both loads (P and W) act at midspan, so:
\[R_A = R_B = \frac{P + W}{2} = \frac{10 + 20}{2} = 15 \text{ kN}\]
(b) Maximum shear force:
Shear force diagram starts at \(V = R_A = 15 \text{ kN}\)
Just before midspan: \(V = 15 - 4 \times 2.5 = 15 - 10 = 5 \text{ kN}\)
Just after midspan: \(V = 5 - 10 = -5 \text{ kN}\)
Maximum shear force: \(|V_{max}| = 15 \text{ kN}\) (at supports)
(c) Maximum bending moment:
Maximum moment occurs at midspan where V = 0:
\[M_{max} = R_A \times 2.5 - \frac{w \times 2.5^2}{2}\]
\[M_{max} = 15 \times 2.5 - \frac{4 \times 6.25}{2} = 37.5 - 12.5 = 25 \text{ kN·m}\]
Alternative: \(M_{max} = \frac{PL}{4} + \frac{wL^2}{8} = \frac{10 \times 5}{4} + \frac{4 \times 25}{8} = 12.5 + 12.5 = 25 \text{ kN·m}\)
Answer: (a) R_A = R_B = 15 kN; (b) |V_max| = 15 kN (at the supports); (c) M_max = 25 kN·m at midspan.