Shear, moment and the sign convention
Cut the beam at any section. The shear V is the sum of the vertical forces on one side of the cut. The bending moment M is the sum of the moments of those forces about the cut. Using the left side of the cut:
Shear at section x
Moment at section x
Uniform load w over length L
Triangular load, 0 to w
Trapezoidal load, w₁ to w₂
Relationships between load, shear and moment
Slope of the shear diagram
Change in shear = −(area of the load diagram)
Slope of the moment diagram
Change in moment = area of the shear diagram
| Load on the segment | Shear diagram | Moment diagram |
|---|---|---|
| No load | Horizontal line | Straight sloping line |
| Uniform load | Sloping straight line | Parabola (2nd degree) |
| Triangular or trapezoidal load | Parabola (2nd degree) | Cubic curve (3rd degree) |
| Concentrated load P | Sudden jump of P | Sharp corner (change of slope) |
| Applied moment (couple) M | No change | Sudden jump of M (up for clockwise) |
Deflection and fixed supports
Loads bend the beam into a curve. How far the beam moves down (the deflection y) comes from the moment diagram: integrate it twice. The answer comes out as a number divided by EI, where E is the modulus of elasticity of the material and I is the moment of inertia of the section. A stiffer beam (bigger EI) deflects less.
Double integration
Slope
Deflection
The constants and come from what the supports allow:
| Support | What it stops | Reactions | Condition to use |
|---|---|---|---|
| Pin or roller | Moving up or down | R | |
| Fixed end | Moving and rotating | R and a moment M | |
| Free end | Nothing | none |
Deflection formulas worth memorizing
| Beam and load | Maximum deflection | Where |
|---|---|---|
| Simple beam, P at midspan | Midspan | |
| Simple beam, uniform w | Midspan | |
| Cantilever, P at the free end | Free end | |
| Cantilever, uniform w | Free end | |
| Cantilever, moment M at the free end | Free end | |
| Propped cantilever, uniform w | (roller reaction ) | 0.578L from the fixed end |
| Both ends fixed, uniform w | (end moments ) | Midspan |
Beam lab
Drag the supports and loads on the beam. Drag the small round handles to stretch a distributed load. You can also type exact values in the list below the diagrams.
Tip: tap a load to change or remove it. Drag a load, or a pin or roller, along the beam. Positions snap to 0.1 m. A fixed support stays at its end of the beam.
Loads on the beam
Loads act downward. Use a negative P for an upward load. Distributed loads go from w₁ (left end) to w₂ (right end): equal values make a uniform load, a zero at one end makes a triangle.
Reactions
Maximum values
Deflection
Type EI to turn the deflection into millimetres.
Shear and moment at key points
| x (m) | V left (kN) | V right (kN) | M (kN·m) | EI·y (kN·m³) |
|---|
Worked examples
A simply supported beam 8 m long carries a uniform load of 10 kN/m over its whole length and a concentrated load of 20 kN at 3 m from the left support A. Find the reactions and the maximum moment.
Show solutionHide solution
Shear just right of the 20 kN load: . It reaches zero at .
R_A = 52.5 kN, R_B = 47.5 kN, M max = 112.81 kN·m at x = 3.25 mA beam 8 m long rests on a pin at A (x = 0) and a roller at B (x = 6 m). It carries a uniform load of 12 kN/m from A to B and a 15 kN load at the free end (x = 8 m). Find the reactions, the maximum positive moment and the moment at B.
Show solutionHide solution
Zero shear at :
R_A = 31 kN, R_B = 56 kN, M+ = 40.04 kN·m, M at B = −30 kN·mA simply supported beam 8 m long carries a load that increases uniformly from 0 at A to 18 kN/m at B. Find the reactions, where the shear is zero, and the maximum moment.
Show solutionHide solution
The load at x is , so the shear is .
R_A = 24 kN, R_B = 48 kN, V = 0 at 4.62 m, M max = 73.90 kN·mA cantilever 4 m long is fixed at A (x = 0). It carries a uniform load of 6 kN/m over its whole length and a 10 kN load at the free end. Find the reactions at the wall and the deflection of the free end in terms of EI.
Show solutionHide solution
Add the two standard cases (superposition):
R_A = 34 kN, M_A = −88 kN·m, δ = 405.33/EI at the free endA beam 8 m long is fixed at A (x = 0) and rests on a roller at B (x = 8 m). It carries a uniform load of 10 kN/m over the whole span. Find the reactions and the maximum positive moment.
Show solutionHide solution
There are 3 unknowns (R_A, M_A, R_B) but only 2 useful equilibrium equations, so use deflection. Remove the roller: the cantilever end would drop by . The roller pushes it back up by . The end does not move, so:
Zero shear at :
R_A = 50 kN, M_A = −80 kN·m, R_B = 30 kN, M+ = 45 kN·m at x = 5 mQuiz: test yourself
Pick an answer to check it right away. Open the solution when you want to see the steps. You can also build each beam in the beam lab to check.
Want the full review?
This is a sample of what you learn at GERTC. Pre-register for the September 2027 or March 2028 CE Board Exam and join the raffle.