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6.5 · Relate distributed load, shear, and bending moment

Learn to relate distributed load, shear, and bending moment through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Internal Forces and Beam Diagrams

ENGG 130 Engineering Mechanics: Statics — Study topic 6.5

A distributed load acts over a length of a beam, rather than at just one point. To find the internal effects at a chosen section, first balance the full beam’s external loads and reactions, then isolate one side of a cut. The relationships between load, shear, and bending moment let us track how those internal quantities vary with position. This lesson uses planar, statically determinate beams and introductory integration.

What you will learn

  • Explain how a distributed load changes shear along a beam and how shear changes bending moment.
  • Use equilibrium and load–shear–moment relations to find reactions, internal shear, and internal moment.
  • Build piecewise expressions and check values at beam ends and load boundaries.

1. Define the beam, load, and sign convention

Model the beam as a straight, planar body. Let xx measure distance from its left end. A distributed load intensity w(x)w(x) is force per length, such as kN/m. Its resultant over an interval is the signed area under the load-intensity graph, and its line of action passes through the centroid of that area.
For this lesson, take downward load intensity as positive, upward forces as positive in force balance, and counterclockwise moments as positive in moment balance. For internal quantities, positive shear acts upward on the left face of a cut, and positive bending moment is sagging. These choices lead to the relations below. A negative internal result means the actual direction or sense is opposite to the assumed positive one.
dVdx=−w(x),dMdx=V(x)\frac{dV}{dx}=-w(x),\quad \frac{dM}{dx}=V(x)
  • Load intensity has units of force per length; the resultant load has units of force.
  • Keep the load, shear, and moment sign conventions consistent.

2. Relate load, shear, and moment

Imagine cutting out a short beam segment of length Δx\Delta x. Its distributed load is approximately w(x)Δxw(x)\Delta x. Vertical force equilibrium says that the shear at the right end differs from the shear at the left end by the negative of this downward load. Divide by Δx\Delta x and let the segment shrink to obtain the shear relation.
Moment equilibrium on the short segment shows that the change in bending moment is approximately the shear times the segment length. Dividing by Δx\Delta x and taking the limit gives the moment relation. These are local equilibrium statements; support reactions and boundary values still need to be found.
Integrating the relations gives useful area rules: the signed area under the load diagram is the change in shear, and the signed area under the shear diagram is the change in moment. A constant distributed load gives a straight-line shear graph and a curved, quadratic moment graph. With no distributed load, shear is constant and moment varies linearly. A concentrated force causes a jump in shear; an applied couple causes a jump in moment.
V(b)−V(a)=−∫abw(x) dx,M(b)−M(a)=∫abV(x) dxV(b)-V(a)=-\int_a^b w(x)\,dx,\quad M(b)-M(a)=\int_a^b V(x)\,dx
  • A downward distributed load reduces shear as xx increases under this convention.
  • Where shear is zero, moment has a local maximum or minimum if the point lies within a smooth interval.

3. Find reactions, then internal quantities

Begin with the complete beam as the system. Draw its supports and external loads. Replace each distributed load by its resultant to find the support reactions using planar equilibrium. For a statically determinate beam, force and moment balance are enough to find these reactions.
Next, cut the beam at a general position xx. Isolate one side and show the internal shear and moment using the chosen convention. Equilibrium of that piece gives the internal quantities. Alternatively, integrate the load–shear–moment relations from a known boundary value. A free end with no force or couple has zero internal shear and moment at the end.
If the load changes formula or begins or ends, write separate expressions for each interval. Shear and moment remain continuous at a load boundary unless a concentrated force or couple is applied there. Check units throughout: ww in kN/m, VV in kN, and MM in kN·m. Finish by checking whole-beam force and moment balance and the internal values at interval boundaries.
∑Fy=0,∑M=0\sum F_y=0,\quad \sum M=0
  • Use whole-beam equilibrium for reactions, then a cut or integration for internal quantities.
  • Boundary values and continuity help anchor and check the expressions.

Worked example

Uniform load on a simply supported beam

A 6 m beam is simply supported at both ends and carries a uniform downward load of 2 kN/m over its full length. Find the reactions and the shear and moment functions, then evaluate them at midspan.
  1. Find the reactions
    Use the complete beam as the system. The load resultant is 2(6)=122(6)=12 kN at midspan. Moment balance about A gives the reaction at B; vertical force balance then gives the reaction at A.
    By(6)−12(3)=0,Ay+By−12=0B_y(6)-12(3)=0,\quad A_y+B_y-12=0
  2. Write the internal functions
    Cut at distance xx from A and use the left portion. Its distributed load is 2x2x kN, acting at x/2x/2. Vertical force and moment balance give the internal shear and sagging-positive moment.
    Ay=By=6 kN,V(x)=6−2x,M(x)=6x−x2A_y=B_y=6\,\mathrm{kN},\quad V(x)=6-2x,\quad M(x)=6x-x^2
  3. Evaluate and verify
    At midspan, shear is zero and moment is 9 kN·m. At the right support, the expressions give zero moment and shear just before the support reaction. The two reactions total 12 kN, and their moments about A balance the load resultant.
    V(3)=0 kN,M(3)=9 kN m,M(6)=0 kN mV(3)=0\,\mathrm{kN},\quad M(3)=9\,\mathrm{kN\,m},\quad M(6)=0\,\mathrm{kN\,m}
Answer: The reactions are Ay=By=6A_y=B_y=6 kN. For 0<x<60<x<6 m, V(x)=6−2xV(x)=6-2x kN and M(x)=6x−x2M(x)=6x-x^2 kN·m. At midspan, V=0V=0 kN and M=9M=9 kN·m.
Check: Force balance: 6+6−12=06+6-12=0 kN. Moment balance about A: 6(6)−12(3)=06(6)-12(3)=0 kN·m. The moment is zero at both supports.

Worked example

Triangular load on a cantilever

A 4 m cantilever is fixed at its left end and free at its right end. A downward triangular load increases linearly from zero at the fixed end to 3 kN/m at the free end. Find the fixed-end reactions and internal shear and moment as functions of position.
  1. Find the resultant and reactions
    The triangular load area is 12(4)(3)=6\frac12(4)(3)=6 kN. Its line of action is 23(4)\frac23(4) m from A. Whole-beam equilibrium gives the upward fixed-end force and counterclockwise fixed-end couple.
    Ay=6 kN,MA=16 kN mA_y=6\,\mathrm{kN},\quad M_A=16\,\mathrm{kN\,m}
  2. Express the load and shear
    The downward-positive load varies linearly, so w(x)=0.75xw(x)=0.75x kN/m. At the free end both internal shear and moment are zero. Integrate inward from that boundary to find the shear.
    w(x)=0.75x,V(x)=∫x4w(s) ds=6−0.375x2w(x)=0.75x,\quad V(x)=\int_x^4 w(s)\,ds=6-0.375x^2
  3. Find moment and check
    Integrating the shear relation from the free end gives a negative moment for this cantilever under downward loading. At the fixed end its magnitude matches the external fixed-end couple. Whole-beam force and moment balance also hold.
    M(x)=−∫x4V(s) ds=−16+6x−0.125x3M(x)=-\int_x^4 V(s)\,ds=-16+6x-0.125x^3
Answer: The fixed-end reactions are 6 kN upward and 16 kN·m counterclockwise. For 0≤x≤40\le x\le4 m, V(x)=6−0.375x2V(x)=6-0.375x^2 kN and M(x)=−16+6x−0.125x3M(x)=-16+6x-0.125x^3 kN·m.
Check: The load resultant is 6 kN, balancing the upward reaction. Its clockwise moment about A is 6(8/3)=166(8/3)=16 kN·m, balanced by the fixed-end couple. At x=4x=4 m, both internal quantities are zero.

Worked example

Load over part of a simply supported beam

A 5 m simply supported beam carries a uniform downward load of 2 kN/m only from x=1x=1 m to x=4x=4 m, measured from the left support. Find shear and moment in each interval.
  1. Solve the reactions
    The load resultant is 2(3)=62(3)=6 kN at x=2.5x=2.5 m. Moment and vertical force balance give 3 kN upward at each support.
    Ay=By=3 kNA_y=B_y=3\,\mathrm{kN}
  2. Build the piecewise shear
    Before the load begins, shear is the left reaction. Across the loaded interval, shear decreases by 2 kN for every metre. After the load ends, shear stays constant up to the right support.
    V(x)={3,0<x<13−2(x−1),1<x<4−3,4<x<5 kNV(x)=\begin{cases}3,&0<x<1\\3-2(x-1),&1<x<4\\-3,&4<x<5\end{cases}\,\mathrm{kN}
  3. Build moment and check continuity
    Moment begins at zero at A and is the integral of shear. There is no applied couple at either load boundary, so moment is continuous there. The adjacent expressions both give 3 kN·m at x=1x=1 m and at x=4x=4 m.
    M(x)={3x,0≤x≤13x−(x−1)2,1≤x≤415−3x,4≤x≤5 kN mM(x)=\begin{cases}3x,&0\le x\le1\\3x-(x-1)^2,&1\le x\le4\\15-3x,&4\le x\le5\end{cases}\,\mathrm{kN\,m}
Answer: The reactions are 3 kN upward at each support. The piecewise functions apply on their stated intervals; moment is continuous and equals 3 kN·m at each end of the loaded region.
Check: The reactions total 6 kN, matching the load. Their moments about A total 15 kN·m, matching the 6 kN resultant at 2.5 m. The moment expression is zero at both supports.

Common mistakes and how to avoid them

Treating load intensity as if it were a force.
Correction: Multiply or integrate intensity over a length to obtain force, and retain the correct units.
Using the same sign for shear change and a downward-positive load.
Correction: With this lesson’s convention, dV/dx=−wdV/dx=-w.
Assuming moment is zero wherever distributed load is zero.
Correction: Zero distributed load means shear is constant; moment can still vary linearly.
Forgetting to split the beam where the load begins or ends.
Correction: Write expressions interval by interval and check values at each boundary.

Lesson summary

  • Distributed load changes shear according to dV/dx=−wdV/dx=-w under the stated convention.
  • Shear changes bending moment according to dM/dx=VdM/dx=V.
  • Areas under the load and shear diagrams give changes in shear and moment, respectively.
  • Find reactions by whole-beam equilibrium, then use cuts or integration and verify the results.

Check your understanding

Question 1

A constant downward load of 4 kN/m acts over a 2 m interval. What is the change in shear across that interval using this lesson’s convention?
  1. An increase of 8 kN
  2. A decrease of 8 kN
  3. No change
  4. A decrease of 2 kN
Show answer and explanation
A decrease of 8 kN
The signed change is the negative area under the downward-positive load: −4(2)=−8-4(2)=-8 kN.

Question 2

On an interval with no distributed load, what is the shape of the bending-moment graph?
  1. A straight line, because shear is constant
  2. A parabola, because shear is constant
  3. A horizontal line in every case
  4. A vertical jump
Show answer and explanation
A straight line, because shear is constant
With w=0w=0, shear is constant; since dM/dx=VdM/dx=V, moment varies linearly with position.

Question 3

For the beam in the first worked example, what is the bending moment at x=3x=3 m?
  1. 0 kN·m
  2. 6 kN·m
  3. 9 kN·m
  4. 18 kN·m
Show answer and explanation
9 kN·m
Substitution into M(x)=6x−x2M(x)=6x-x^2 gives M(3)=18−9=9M(3)=18-9=9 kN·m.

Key terms

Distributed load intensity
Force applied per unit length of a beam, commonly measured in kN/m.
Shear, VV
The internal transverse force at a beam section.
Bending moment, MM
The internal moment at a beam section, reported here with sagging taken as positive.
Resultant load
A single force equal to the total distributed load, applied at the centroid of its load-intensity area.

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