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6.3 · Construct axial-force and shear-force diagrams

Learn to construct axial-force and shear-force 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 tracking internal forces along a beam or member

An axial-force diagram and a shear-force diagram show how internal forces vary from one end of a member to the other. To construct them, first find the external reactions, then imagine cutting the member at a position x and isolate one side of the cut. Equilibrium of that part gives the internal force at the cut. Repeating this reasoning across the member produces the diagrams. The examples use planar, statically determinate members and SI units.

What you will learn

  • Explain what axial-force and shear-force diagrams show.
  • Use a consistent sign convention to calculate internal axial force and shear at a section.
  • Find support reactions before constructing diagrams.
  • Use point loads and distributed loads to identify diagram jumps and slopes.
  • Check a completed diagram against force and moment equilibrium.

1. What the diagrams represent

The system is the beam or straight member being studied. At a cut through it, the material on one side exerts forces on the other side. In this lesson we track two components: axial force N, directed along the member, and shear force V, directed perpendicular to it. A diagram plots each quantity against the position x along the member.
Use the following sign convention throughout: positive N is tension; negative N is compression. Positive V is defined as the algebraic sum of vertical external forces acting to the left of the cut. Thus a positive upward reaction to the left contributes positively, and a downward load to the left contributes negatively. This definition makes the shear diagram easy to calculate from the left portion; the equal-and-opposite force on that isolated portion balances this sum.
A point force causes an abrupt change in the relevant diagram as the cut passes that force. A distributed load causes shear to change continuously. Between loads that affect a quantity, its diagram is constant. Mark the member ends, load locations, values, and units so each section of the diagram is clear.
∑Fx=0,∑Fy=0,∑M=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M=0
  • Use N for axial force and V for shear force.
  • State the sign convention before calculating values.
  • A diagram's horizontal coordinate is position along the member.

2. Reactions, cuts, and diagram rules

Begin with a free-body diagram of the whole member. Replace supports by their allowed reaction components, show applied loads and distances, and solve the equilibrium equations. A pin can exert horizontal and vertical reactions; a roller on a horizontal surface exerts a vertical reaction. Once reactions are known, choose an arbitrary cut at x and isolate the left part.
For the left part, apply force equilibrium. With the convention used here, V(x) is the sum of the vertical external forces to the left of the cut. For axial force, add the signed horizontal external forces to the left; a positive result represents tension under the stated convention. If an answer is negative, the actual internal force is opposite to the assumed positive direction.
For a downward distributed load with intensity q(x), measured in force per length, the shear diagram decreases as x increases. Over an interval, the change in shear equals the negative area under q. For a constant load, this gives a straight sloping segment; for a linearly varying load, it gives a curved segment. A concentrated vertical force produces a shear jump equal to the negative of that force when the force points downward.
A useful construction order is: mark the reaction and load positions, calculate shear values just to either side of each point load, then connect values according to any distributed loading. Construct the axial-force diagram in the same way using axial loads. Check that the diagram returns to zero at a free end with no unbalanced end force.
V(x2)−V(x1)=−∫x1x2q(x) dxV(x_2)-V(x_1)=-\int_{x_1}^{x_2}q(x)\,dx
  • Solve whole-member reactions before making internal cuts.
  • For downward q, shear change over an interval is the negative load area.
  • Point forces create jumps; distributed forces create continuous changes.

3. Reading and checking the diagrams

At any position, the diagram's ordinate gives the signed internal force. A constant segment means there is no distributed load affecting that force over the interval. A jump identifies a concentrated force; its size and direction should match the applied load. A sloping or curved shear segment should correspond to the area of the applied distributed load.
Use equilibrium as an independent check. For the whole member, the support reactions and applied loads must sum to zero in both coordinate directions, and their moments about a convenient point must sum to zero. For the shear diagram, account for every jump and every change due to distributed loading; the value at the far end must agree with the end-force balance. For axial force, likewise account for each axial load and check the end balance.
Keep units visible: axial force and shear are measured in N or kN, while a distributed-load intensity is measured in N/m or kN/m. Do not label a force diagram with moment units.
[N]=[V]=kN,[q]=kN/m[N]=[V]=\mathrm{kN},\quad [q]=\mathrm{kN/m}
  • Compare each diagram feature with its corresponding load.
  • Verify whole-member force and moment equilibrium.
  • Keep force and distributed-load units distinct.

Worked example

1. Constant axial force in a member

A straight horizontal member is fixed at A and pulled leftward at its right end B by 12 kN. Construct its axial-force diagram. The member's length is 3 m, and there are no vertical loads.
  1. Define the system and find the reaction
    Take the whole member as the system. Choose positive x to the right. Horizontal equilibrium requires the fixed-end reaction to balance the leftward 12 kN force. Vertical equilibrium is satisfied with no vertical forces. The force lines lie along the member, so their moments about A are zero.
    Ax−12 kN=0A_x-12\,\mathrm{kN}=0
  2. Cut the member
    Make a cut anywhere between A and B and isolate the left part. The only horizontal external force on that part is the 12 kN reaction at A. By the convention that positive axial force is tension, the internal axial force is 12 kN in tension at every cut.
    N(x)=12 kN,0<x<3 mN(x)=12\,\mathrm{kN},\quad 0<x<3\,\mathrm{m}
  3. Check the member
    The horizontal forces on the full member balance, and the collinear forces create no net moment. The axial diagram is a constant positive value through the member; the end load is balanced at B.
    ∑Fx=12−12=0,∑MA=0\sum F_x=12-12=0,\quad \sum M_A=0
Answer: The axial-force diagram is constant at +12 kN (tension) from A to B.
Check: The reaction and applied force are equal and opposite. Their moment about A is zero because both act along the member's axis.

Worked example

2. Shear changes at concentrated loads

A 6 m simply supported beam has a pin at A and a roller at B. A 10 kN downward point load acts 2 m from A, and a 6 kN downward point load acts 5 m from A. Construct the shear-force diagram.
  1. Find support reactions
    Take moments about A to find the vertical reaction at B. Then use vertical force equilibrium to find the reaction at A. There are no horizontal loads, so the horizontal reaction at the pin is zero.
    6By−10(2)−6(5)=0,Ay+By=166B_y-10(2)-6(5)=0,\quad A_y+B_y=16
  2. Evaluate the reaction values
    Moment balance gives the roller reaction as 8.33 kN upward. The remaining upward reaction at A is 7.67 kN. Keep enough decimal places in intermediate calculations to avoid rounding errors.
    By=8.33 kN,Ay=7.67 kNB_y=8.33\,\mathrm{kN},\quad A_y=7.67\,\mathrm{kN}
  3. Construct the shear segments
    Starting just right of A, the shear is +7.67 kN. Crossing the 10 kN downward force reduces it by 10 kN, giving −2.33 kN. Crossing the 6 kN downward force reduces it again to −8.33 kN. The upward reaction at B brings the final value to zero.
    V={7.67 kN,0<x<2−2.33 kN,2<x<5−8.33 kN,5<x<6V=\begin{cases}7.67\,\mathrm{kN},&0<x<2\\-2.33\,\mathrm{kN},&2<x<5\\-8.33\,\mathrm{kN},&5<x<6\end{cases}
  4. Check force and moment balance
    The upward reactions total 16 kN, matching the downward loads. Taking moments about A gives 8.33 kN times 6 m, equal to the combined load moment of 50 kN·m. The shear value returns to zero after the reaction at B.
    7.67+8.33−10−6=0,8.33(6)−10(2)−6(5)=07.67+8.33-10-6=0,\quad 8.33(6)-10(2)-6(5)=0
Answer: The shear diagram is +7.67 kN from A to the 10 kN load, −2.33 kN between the loads, and −8.33 kN from the 6 kN load to B. It jumps to zero at B.
Check: The jump at each downward point load is negative and equals that load's magnitude. Reaction and load balances are satisfied.

Worked example

3. Shear under a triangular distributed load

A 4 m cantilever is fixed at A. A downward triangular distributed load increases linearly from zero at A to 12 kN/m at its free end B. Construct its shear diagram.
  1. Replace the distributed load by its resultant
    The load is a triangle, so its total is its area: 24 kN downward. Its line of action is two-thirds of the span from the zero-intensity end A, or 8/3 m from A. Whole-member equilibrium therefore gives an upward reaction of 24 kN at A and a counterclockwise fixed-end reaction moment of 64 kN·m.
    R=12(4)(12)=24 kN,MA=24(83)=64 kN ⁣⋅ ⁣mR=\tfrac12(4)(12)=24\,\mathrm{kN},\quad M_A=24\left(\tfrac83\right)=64\,\mathrm{kN\!\cdot\!m}
  2. Write the load intensity along the beam
    Let x be measured in metres from A. Since intensity rises linearly from 0 to 12 kN/m over 4 m, its value at x is 3x kN/m downward. The shear at a cut is the upward reaction minus the total downward load to the left of that cut.
    q(x)=3x kN/m,V(x)=24−∫0x3s dsq(x)=3x\,\mathrm{kN/m},\quad V(x)=24-\int_0^x 3s\,ds
  3. Evaluate and interpret the shear
    Integrating gives a curved, parabolic shear diagram. It starts at +24 kN just to the right of A and decreases continuously to zero at the free end. There is no concentrated force at B, so there is no jump there.
    V(x)=24−1.5x2 kN,0<x<4 mV(x)=24-1.5x^2\,\mathrm{kN},\quad 0<x<4\,\mathrm{m}
  4. Verify equilibrium
    The reaction balances the 24 kN resultant. The 64 kN·m support moment balances the clockwise moment of the resultant about A. The zero shear at B agrees with the absence of an end force.
    24−24=0,64−24(83)=024-24=0,\quad 64-24\left(\tfrac83\right)=0
Answer: The shear diagram is V(x)=24−1.5x2V(x)=24-1.5x^2 kN for positions between A and B. It falls smoothly from +24 kN to 0 kN.
Check: The area under the downward load is 24 kN, exactly the upward support reaction. The resultant's moment is balanced by the fixed-end reaction moment.

Common mistakes and how to avoid them

Changing the shear sign convention halfway through the calculation.
Correction: Write the convention at the start and use the signed external vertical forces to the left of each cut.
Drawing a jump in shear under a distributed load.
Correction: A distributed load changes shear continuously; its accumulated area sets the change. A concentrated force creates a jump.
Using the load value at a point as the shear change across an interval.
Correction: Use the total force over the interval, which is the area under the distributed-load intensity.
Stopping after finding the support reactions.
Correction: Calculate internal values across each region, mark jumps and slopes, and check the end balance.
Treating a negative result as automatically incorrect.
Correction: A negative result means the actual internal force acts opposite to the direction defined as positive.

Lesson summary

  • Find support reactions from equilibrium of the complete member.
  • Use a cut and equilibrium of one side to determine internal axial force and shear.
  • Positive axial force is tension; this lesson defines positive shear as the signed vertical external force to the left of the cut.
  • Point loads cause diagram jumps, while distributed loads change shear by their accumulated area.
  • Check force and moment balance and confirm the diagram values agree with the loading.

Check your understanding

Question 1

A downward 4 kN point load is crossed while moving from left to right along a beam. Under the stated shear convention, what is the jump in V?
  1. A decrease of 4 kN
  2. An increase of 4 kN
  3. No change
  4. A change of 4 kN·m
Show answer and explanation
A decrease of 4 kN
A downward point load contributes negatively to the sum of vertical forces to the left of the cut, so shear drops by 4 kN.

Question 2

A beam carries a uniform downward load of 3 kN/m over a 2 m interval. What is the change in shear from the beginning to the end of that interval?
  1. A decrease of 6 kN
  2. An increase of 6 kN
  3. A decrease of 3 kN
  4. No change
Show answer and explanation
A decrease of 6 kN
The downward load over the interval is its intensity times length, or 6 kN. Its sign makes the shear change −6 kN.

Question 3

A calculated axial force is −5 kN using tension as positive. What does this indicate?
  1. The member is in compression with magnitude 5 kN.
  2. The member is in tension with magnitude 5 kN.
  3. The force should be reported as 5 kN·m.
  4. The member has no axial force.
Show answer and explanation
The member is in compression with magnitude 5 kN.
A negative axial result is opposite the assumed tensile direction, so it represents 5 kN in compression.

Key terms

Axial force
Internal force acting along the length of a member.
Shear force
Internal force acting perpendicular to the member's length in the plane considered.
Distributed load
A load spread along a length, described by force per unit length.
Resultant
A single force equal to the total effect of a distributed load, acting at its line of action.
Free-body diagram
A drawing of an isolated body with its external loads and support reactions shown.

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