DoAssignment.ca

6.4 · Construct bending-moment diagrams

Learn to construct bending-moment diagrams through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Internal Forces and Beam Diagrams

A first-year statics method for finding and sketching internal bending moment along a beam

A bending-moment diagram shows how the internal bending moment varies from one end of a beam to the other. To construct one, first find the beam’s support reactions. Then work along the beam, tracking how the loads change the internal shear force and bending moment. The beam is treated as a rigid body in planar statics, and the internal quantities are found by considering equilibrium of a cut portion. In this lesson, positive bending moment means sagging: the beam’s usual positive-curvature shape. All examples use kN and m, so bending moments are in kN·m.

What you will learn

  • Use whole-beam equilibrium to find support reactions for a statically determinate beam.
  • Relate applied loads, shear force, and bending moment as you move along a beam.
  • Construct a bending-moment diagram by identifying its values, shape, and important changes.
  • Check a diagram using units, boundary conditions, and equilibrium.

1. Start with the complete beam and its free-body diagram

The system is the entire beam, isolated from its supports and loads. Draw its free-body diagram with all applied forces, applied couples, support locations, and dimensions. For a pin support, show horizontal and vertical reaction components; for a roller on a horizontal surface, show a vertical reaction. If there is no horizontal applied load, horizontal equilibrium will usually make the pin’s horizontal reaction zero.
Choose a coordinate xx measured from the left end of the beam. Take upward forces as positive and counterclockwise moments as positive when finding reactions. Apply equilibrium to the whole beam before making any cut. For a beam with vertical loading, the useful equations are force balance vertically and moment balance about a convenient point. Include the horizontal equation if horizontal forces are present.
Once reactions are known, make an imaginary cut at a general position xx and isolate one side. The cut exposes internal shear force VV and internal bending moment MM. They are the force and moment that the removed part of the beam would exert on the isolated part. Use a consistent sign convention: positive bending moment is sagging. A positive shear convention can be used consistently on the cut; the examples below identify the shear values from the change in moment.
∑Fy=0,∑M=0\sum F_y=0,\quad \sum M=0
  • Find reactions from the complete-beam free-body diagram before constructing the diagram.
  • Use distances from a clearly stated origin when taking moments.
  • Keep force units and moment units distinct.

2. Track how loads shape the diagram

The key relationship for constructing a bending-moment diagram is that its slope is the shear force. Between concentrated forces, a constant shear gives a straight-line moment diagram. A downward distributed load makes shear decrease as you move right; consequently, the moment diagram curves. For a uniform downward load, shear changes linearly and moment changes quadratically.
A concentrated vertical force causes a sudden change in shear equal to the force, with sign set by its direction. The bending moment remains continuous at that point if no applied couple is there. A concentrated applied couple causes a sudden change in bending moment; it does not act like a vertical force. At an unloaded end, or at a simple support at the end of a beam with no applied end couple, the bending moment is zero.
A reliable sketch begins by marking beam endpoints, supports, point loads, distributed-load limits, and any applied couples. Calculate moment values at those locations, and note where shear is zero within a region. A change of shear from positive to negative indicates a local maximum of moment; a change from negative to positive indicates a local minimum. Then connect the values with the correct straight or curved shape.
dMdx=V,dVdx=−w\frac{dM}{dx}=V,\quad \frac{dV}{dx}=-w
  • Constant shear produces a straight-line bending-moment segment.
  • A point force changes shear; a point couple changes bending moment.
  • Where shear is zero, the bending-moment diagram may have a local extreme.

3. A repeatable construction and checking method

After drawing the free-body diagram, solve the support reactions with whole-beam equilibrium. Divide the beam into regions wherever a point force or distributed load begins or ends. For each region, cut at a general position xx and write moment equilibrium for one side of the cut. This produces an expression for M(x)M(x) in that region. Keeping the expressions separate avoids applying a load before the cut has passed its location.
Substitute the region endpoints into each expression and plot the resulting moment values at their correct positions. Use the shear on each region to determine whether the moment rises or falls. Check that neighboring expressions give the same moment at a point-force location when no couple is applied there. If there is a couple, check for the corresponding jump instead.
Finally, verify that the original reactions satisfy both vertical force balance and moment balance. Check that the bending-moment diagram has the expected end values, that its slopes agree with the shear signs, and that its units are kN·m. A negative value is not automatically an error: it means the actual bending sense is opposite to the positive sagging convention.
M(x)=∫V(x) dxM(x)=\int V(x)\,dx
  • Write one moment expression for each region separated by a load change.
  • Check diagram continuity at point forces and account for jumps at couples.
  • Verify reactions independently using force and moment equilibrium.

4. Reading the completed diagram

The horizontal axis of a bending-moment diagram is the beam position xx; the vertical ordinate is the signed moment MM. Label important values directly at supports, point loads, load boundaries, and internal extrema. A diagram is not defined only by its peak: its sign, shape, and key values also matter.
Use the relation between slope and shear as a check on the sketch. If shear is positive over an interval, moment increases there; if shear is negative, moment decreases. For a distributed load, the change in moment across a region equals the signed area under the shear diagram. This area check is especially useful when calculating an endpoint value without redoing the cut equilibrium.
The examples use simple supports and known loads so the reactions and internal moments can be found from statics alone. Each diagram is therefore based on equilibrium, not on assumptions about how a beam changes shape.
M(x2)−M(x1)=∫x1x2V(x) dxM(x_2)-M(x_1)=\int_{x_1}^{x_2}V(x)\,dx
  • Label signed moment values and units on the vertical axis.
  • The area under the shear diagram gives the change in bending moment.
  • Use statics and load-to-moment relationships, not deformation information.

Worked example

One point load on a simply supported beam

A 6 m beam is supported by a pin at A and a roller at B. A 12 kN downward point load acts 2 m from A. Find the reactions and construct the bending-moment diagram.
  1. Find the reactions
    Take moments about A, with counterclockwise positive. The load is 2 m from A and B is 6 m from A. Vertical force balance then gives the reaction at A.
    6By−12(2)=0,Ay+By−12=06B_y-12(2)=0,\quad A_y+B_y-12=0
  2. Write the moment by region
    Cut the beam at position xx and use the left portion. Before the point load, only the reaction at A contributes. After the load, its moment contribution must also be included.
    M(x)={8x,0≤x≤28x−12(x−2),2≤x≤6M(x)=\begin{cases}8x,&0\le x\le2\\8x-12(x-2),&2\le x\le6\end{cases}
  3. Locate and check the key values
    The moment rises linearly to the point load and falls linearly afterward. The shear changes from positive to negative at the load, so the maximum moment occurs there. The reactions sum to the applied load, and their moments about A balance the load’s moment.
    Ay=8 kN,By=4 kN,M(2)=16 kN⋅m,M(6)=0A_y=8\,\text{kN},\quad B_y=4\,\text{kN},\quad M(2)=16\,\text{kN}\cdot\text{m},\quad M(6)=0
Answer: The reactions are Ay=8A_y=8 kN upward and By=4B_y=4 kN upward. The bending moment is 8x8x kN·m from A to the load and 24−4x24-4x kN·m from the load to B. The diagram is triangular, with a maximum sagging moment of 16 kN·m at x=2x=2 m.
Check: 8+4−12=08+4-12=0 kN and 4(6)−12(2)=04(6)-12(2)=0 kN·m.

Worked example

Uniformly distributed load over the span

A simply supported beam has a span of 8 m and carries a uniform downward load of 3 kN/m across its full length. Find the reactions and construct the bending-moment diagram.
  1. Replace the load for whole-beam equilibrium
    The load’s resultant is its intensity times the loaded length, acting at the midpoint because the load is uniform. Symmetry or moment equilibrium gives equal vertical reactions.
    W=3(8)=24 kN,Ay=By=12 kNW=3(8)=24\,\text{kN},\quad A_y=B_y=12\,\text{kN}
  2. Use a cut at position x
    For the left portion, the distributed load over length xx has resultant 3x3x kN at its midpoint, x/2x/2 from A. Moment equilibrium about the cut gives the internal moment.
    M(x)=12x−(3x)x2=12x−1.5x2M(x)=12x-(3x)\frac{x}{2}=12x-1.5x^2
  3. Find the maximum and confirm the ends
    The shear is the slope of the moment expression. Setting it to zero identifies the maximum. At both supports the moment is zero, as expected for this beam with no applied end couples.
    V(x)=12−3x,V=0⇒x=4 m,M(4)=24 kN⋅mV(x)=12-3x,\quad V=0\Rightarrow x=4\,\text{m},\quad M(4)=24\,\text{kN}\cdot\text{m}
Answer: Each support reaction is 12 kN upward. The moment diagram is a downward-opening parabola, M(x)=12x−1.5x2M(x)=12x-1.5x^2 kN·m for 0≤x≤80\le x\le8. It is zero at both supports and reaches 24 kN·m at midspan.
Check: 12+12−24=012+12-24=0 kN and 12(8)−24(4)=012(8)-24(4)=0 kN·m.

Worked example

Two point loads and a shear reversal

A 9 m simply supported beam carries a 6 kN downward load at x=3x=3 m and a 9 kN downward load at x=6x=6 m. Find the reactions and construct the bending-moment diagram.
  1. Solve the support reactions
    Take moments about A. The 6 kN and 9 kN loads act at 3 m and 6 m, respectively. Then use vertical force balance to find the other reaction.
    9By−6(3)−9(6)=0,Ay+By=159B_y-6(3)-9(6)=0,\quad A_y+B_y=15
  2. Construct the three regions
    For each region, include only forces to the left of the cut. Each point load changes the slope by changing the shear, while the moment itself remains continuous at the load.
    M(x)={233x,0≤x≤3233x−6(x−3),3≤x≤6233x−6(x−3)−9(x−6),6≤x≤9M(x)=\begin{cases}\frac{23}{3}x,&0\le x\le3\\\frac{23}{3}x-6(x-3),&3\le x\le6\\\frac{23}{3}x-6(x-3)-9(x-6),&6\le x\le9\end{cases}
  3. Read the values and verify
    The shear is positive before the first load, still positive but smaller between loads, and negative after the second load. Thus the moment keeps rising to the second load, then falls. Evaluate the expressions at the load positions and at B.
    Ay=223 kN,By=233 kN,M(3)=22 kN⋅m,M(6)=23 kN⋅m,M(9)=0A_y=\frac{22}{3}\,\text{kN},\quad B_y=\frac{23}{3}\,\text{kN},\quad M(3)=22\,\text{kN}\cdot\text{m},\quad M(6)=23\,\text{kN}\cdot\text{m},\quad M(9)=0
Answer: The reactions are Ay=22/3A_y=22/3 kN and By=23/3B_y=23/3 kN, both upward. The diagram is linear in each of the three regions, with values 0 at A, 22 kN·m at x=3x=3 m, 23 kN·m at x=6x=6 m, and 0 at B. The maximum is 23 kN·m at the second load, where shear changes from positive to negative.
Check: 223+233−15=0\frac{22}{3}+\frac{23}{3}-15=0 kN and 233(9)−6(3)−9(6)=0\frac{23}{3}(9)-6(3)-9(6)=0 kN·m.

Common mistakes and how to avoid them

Treating a point force as a jump in the bending-moment diagram.
Correction: A point force changes shear. Moment remains continuous there unless an applied couple is also present.
Drawing a curved moment diagram under a constant shear.
Correction: Since the slope of moment is shear, constant shear gives a straight-line moment segment.
Forgetting that a distributed load over a cut portion has a resultant and a location.
Correction: For a uniform load, use the intensity times the portion length, acting at that portion’s midpoint.
Using inconsistent units or mixing the signs of loads and reactions.
Correction: Choose a force and moment sign convention at the start, and carry kN, m, and kN·m consistently.

Lesson summary

  • Draw the whole-beam free-body diagram and solve reactions using equilibrium.
  • Cut the beam into regions wherever loads change and write a moment expression in each region.
  • Use shear as the slope of the bending-moment diagram; point forces change shear, while couples create moment jumps.
  • Mark endpoint and load-location values, identify extrema, sketch the correct shapes, and verify force and moment balance.

Check your understanding

Question 1

A beam segment has constant positive shear. What shape does its bending-moment diagram have on that segment?
  1. A straight line rising to the right
  2. A straight line falling to the right
  3. A horizontal line
  4. A downward-opening parabola
Show answer and explanation
A straight line rising to the right
The moment slope equals shear. Constant positive shear means a constant positive slope, so moment rises linearly.

Question 2

At a point load with no applied couple, what happens to the bending moment as the cut passes the load?
  1. It must jump by the magnitude of the point load.
  2. It remains continuous, although its slope can change.
  3. It becomes zero at the load.
  4. It must change from positive to negative.
Show answer and explanation
It remains continuous, although its slope can change.
The point force changes shear, which changes the slope of the moment diagram. Without an applied couple, the moment itself remains continuous.

Question 3

A simply supported beam has a 4 m span and a 10 kN downward point load at midspan. What is its maximum sagging moment?
  1. 5 kN·m
  2. 10 kN·m
  3. 20 kN·m
  4. 40 kN·m
Show answer and explanation
10 kN·m
The reactions are 5 kN each. At midspan, the moment is 5(2)=105(2)=10 kN·m, and the shear changes from positive to negative there.

Key terms

Bending-moment diagram
A graph showing the signed internal bending moment along a beam.
Shear force
The internal force on a beam cross-section that balances the transverse forces on one side of a cut.
Sagging
The positive bending sense used in this lesson for the moment diagram.
Resultant
A single force that has the same overall effect as a distributed load for equilibrium calculations.

Continue through ENGG 130

View the complete ENGG 130 University of Alberta ENGG 130: Engineering Mechanics: Statics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta ENGG 130: Engineering Mechanics: Statics, study topic 6.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question