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Engineering mechanics: How are forces and reactions distributed through beams, trusses and frames in civil structures?

Apply equilibrium of forces and moments to analyse simply supported beams and pin-jointed trusses, calculate support reactions and internal member forces, and identify members in tension and compression

A focused answer to the HSC Engineering Studies Civil Structures dot point on force analysis. Equilibrium, support reactions on simply supported beams, the method of joints for pin-jointed trusses, the Sydney Harbour Bridge example, and worked past exam questions.

Generated by Claude Opus 4.810 min answerUpdated 2026-07-03

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to apply the two equations of static equilibrium ( $\sum F = 0$ and $\sum M = 0$ ) to simply supported beams under point and distributed loads, and to pin-jointed trusses where every member is in pure tension or pure compression. You must be able to draw a free-body diagram, find support reactions, and use the method of joints to find internal member forces.

The answer

A civil structure in equilibrium has zero net force and zero net moment. Two equations let you solve for two reaction forces on a simply supported beam. For a pin-jointed truss, you first find the support reactions, then work joint by joint.

Equilibrium of a simply supported beam

For any loaded beam in equilibrium:

\sum F_y = 0 \qquad \sum M = 0

Take moments about one support to eliminate its reaction from the equation. Solve for the other reaction, then use vertical equilibrium to find the first.

A uniformly distributed load (UDL) of $w$ N/m over length $L$ is treated as a single point load of magnitude $wL$ at the midpoint of the loaded span.

An owned free-body diagram makes the standard beam case concrete: a point load $P$ at distance $a$ from the left support, plus a UDL of $w$ over the full span $L$ , with the two unknown reactions $R_A$ and $R_B$ .

Method of joints for trusses

Truss members carry axial force only, either tension (pulling away from the joint) or compression (pushing into the joint). At each joint:

\sum F_x = 0 \qquad \sum F_y = 0

Start at a joint with at most two unknown member forces. Assume both unknowns are in tension (positive). A negative answer means the member is in compression.

A Warren truss (the alternating triangle pattern used in many footbridges and roof trusses) shows the pattern clearly: under loading from above, the top chord is squeezed short (compression) and the bottom chord is stretched long (tension), exactly like the two sides of a bent beam.

Shear force and bending moment

Once reactions are known, the internal effects along a beam are described by the shear force (the net vertical force to one side of a section) and the bending moment (the net moment to one side of a section). The standard sign convention takes upward forces on the left as positive shear, and sagging (concave-up) as positive bending. Three rules speed up sketching the diagrams:

A point load causes a step in the shear-force diagram equal to the load.
A uniformly distributed load makes the shear-force diagram slope linearly and the bending-moment diagram curve parabolically.
The bending moment is a maximum where the shear force passes through zero, which is exactly the section the designer checks against the allowable bending stress.

For a simply supported beam with a central point load $P$ over span $L$ , the maximum bending moment is $M_{\max} = PL/4$ ; for a full-length UDL it is $M_{\max} = wL^2/8$ . These two standard cases recur constantly in the HSC paper.

Tension and compression members

In a pin-jointed truss every member is a two-force member: the force acts along the member axis only. A member that is being pulled (force arrows pointing away from the joints, away from the member centre) is in tension; one being pushed (arrows pointing into the joints) is in compression. Tension members can be slender because they cannot buckle; compression members must be stockier to resist buckling, which is why truss top chords (usually in compression) are often heavier sections than the bottom chords. Identifying which members are which lets the designer size each member economically.

Sydney Harbour Bridge as the canonical Australian example

The Sydney Harbour Bridge is a two-hinged steel arch with a deck supported by hangers. The main arch carries compression; the hangers carry tension; the deck carries bending. The bridge's design (Bradfield, 1924) used hand calculations of equilibrium at every joint of the analysis truss before fabrication began. The same equilibrium equations sit behind every modern finite-element civil-engineering package.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC style5 marksA simply supported beam of length 6 m carries a point load of 12 kN at 2 m from the left support and a uniformly distributed load of 4 kN/m over the full length. Calculate the left and right support reactions.

Show worked answer →

Apply equilibrium. Take moments about the left support and sum vertical forces.

Total UDL force: $w \times L = 4 \times 6 = 24$ kN, acting at the midspan ( $x = 3$ m).

Moments about the left support (clockwise positive):

\sum M_L = 0 = (12)(2) + (24)(3) - R_R(6)

R_R = \frac{24 + 72}{6} = 16 \text{ kN}

Sum of vertical forces:

R_L + R_R = 12 + 24 = 36 \text{ kN}

R_L = 36 - 16 = 20 \text{ kN}

Markers reward (1) a clear free-body diagram showing the loads and reactions, (2) the moment equation set out explicitly, (3) the second equation from vertical equilibrium, and (4) both reactions with units. Show the UDL as a single resultant at midspan before taking moments.

HSC 20202 marksState the two conditions of static equilibrium for a coplanar force system and explain why both are needed to analyse a loaded beam.

Show worked answer →

The two conditions are that the net force is zero ( $\sum F = 0$ , applied to both horizontal and vertical components) and that the net moment about any point is zero ( $\sum M = 0$ ). Both are needed because a beam can be balanced for forces yet still rotate (a couple), or balanced for moments yet still translate. Only when both hold is the body truly in equilibrium. Markers award one mark for each correctly stated condition with the second mark depending on the reason that force balance alone does not prevent rotation.

HSC 20236 marksA cantilever beam 3.0 m long is fixed at the left wall. It carries a uniformly distributed load of 6 kN/m over its full length and a point load of 10 kN at the free end. Calculate the vertical reaction and the fixing moment at the wall, and state the location and magnitude of the maximum bending moment.

Show worked answer →

Resultant of the UDL. $W = wL = 6 \times 3.0 = 18$ kN, acting at the midspan, 1.5 m from the wall.

Vertical reaction. Sum vertical forces:

R = W + P = 18 + 10 = 28 \text{ kN (upward)}

Fixing moment. Take moments about the wall (the fixed support):

M_{\text{wall}} = (18)(1.5) + (10)(3.0) = 27 + 30 = 57 \text{ kN m}

Maximum bending moment. On a cantilever the bending moment is greatest at the fixed end, so the maximum bending moment is $57$ kN m at the wall, where the fixing moment resists it. Markers reward the UDL resultant placed at midspan, the reaction from vertical equilibrium, the moment taken about the fixed end, and the identification that the maximum bending moment on a cantilever always occurs at the support.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA simply supported beam of length 5 m carries a single point load of 15 kN located 1 m from the left support. Calculate both support reactions.

Show worked solution →

Take moments about the left support (clockwise positive):

\sum M_L = 0 = (15)(1) - R_R(5)

R_R = \frac{15}{5} = 3 \text{ kN}

Sum of vertical forces:

R_L + R_R = 15 \implies R_L = 15 - 3 = 12 \text{ kN}

Marking criteria: 1 mark for a correct moment equation about a support, 1 mark for the correct value of $R_R$ (3 kN), 1 mark for the correct value of $R_L$ (12 kN) with units.

foundation3 marksAt one joint of a pin-jointed truss, a vertical load of 8 kN acts downward. Only two members meet at this joint: a horizontal member and a member inclined at 45 degrees above the horizontal. Use the method of joints (assume both members are in tension) to find the force in each member, and state whether each is actually in tension or compression.

Show worked solution →

Vertical equilibrium (only the inclined member has a vertical component):

F_{\text{incl}} \sin 45^\circ = 8 \implies F_{\text{incl}} = \frac{8}{\sin 45^\circ} = 11.3 \text{ kN}

The sign is positive, so the inclined member is in tension (11.3 kN).

Horizontal equilibrium:

F_{\text{horiz}} + F_{\text{incl}} \cos 45^\circ = 0 \implies F_{\text{horiz}} = -11.3 \times \cos 45^\circ = -8.0 \text{ kN}

The sign is negative, so the horizontal member is actually in compression (8.0 kN), not tension as assumed.

Marking criteria: 1 mark for the correct vertical equilibrium equation and value (11.3 kN tension), 1 mark for the correct horizontal equilibrium equation and value (8.0 kN), 1 mark for correctly identifying the horizontal member as compression from the negative sign.

core4 marksA simply supported beam 8 m long carries a point load of 20 kN at 3 m from the left support and a uniformly distributed load of 3 kN/m over the full span. Calculate both support reactions, then calculate the bending moment immediately under the point load (you do not need to check whether this is the global maximum).

Show worked solution →

UDL resultant: $W = wL = 3 \times 8 = 24$ kN, acting at midspan (4 m).

Moments about the left support:

\sum M_L = 0 = (20)(3) + (24)(4) - R_R(8)

R_R = \frac{60 + 96}{8} = 19.5 \text{ kN}

Vertical equilibrium:

R_L = 20 + 24 - 19.5 = 24.5 \text{ kN}

Bending moment at $x = 3$ m (sum of moments of everything to the left of the section): the UDL acting over the first 3 m has resultant $3 \times 3 = 9$ kN at 1.5 m from the left support, so its moment arm about $x = 3$ m is $3 - 1.5 = 1.5$ m.

M_{x=3} = R_L(3) - (9)(1.5) = 24.5 \times 3 - 13.5 = 73.5 - 13.5 = 60 \text{ kN m}

Marking criteria: 1 mark for the correct reactions (24.5 kN and 19.5 kN), 1 mark for correctly reducing the partial UDL to a resultant with the correct moment arm, 1 mark for the correct bending moment value (60 kN m) with units, 1 mark for working entirely from one side of the section.

core5 marksThe table below gives the shear force

V(x)

immediately to the left and right of two sections on a simply supported beam of length 6 m, loaded by a single point load. Use the table to state the location of the maximum bending moment and calculate its value. | Section | x = 0 m | x = 2 m (just left) | x = 2 m (just right) | x = 6 m | |---|---|---|---|---| | V (kN) | +14 | +14 | -10 | -10 |

Show worked solution →

Interpreting the table: The shear force is constant at $+14$ kN from $x = 0$ to just left of $x = 2$ m, then jumps abruptly to $-10$ kN just right of $x = 2$ m, and stays constant to $x = 6$ m. A step in the shear-force diagram occurs only where a point load is applied, so the single point load acts at $x = 2$ m and has magnitude $14 - (-10) = 24$ kN.
Locating the maximum bending moment: The bending moment is a maximum where the shear force passes through (or jumps through) zero. Since the diagram is constant either side of $x = 2$ m and only crosses zero at the jump itself, the maximum bending moment occurs at $x = 2$ m, directly under the point load.
Calculating the maximum bending moment: The bending moment at a section equals the area under the shear-force diagram up to that section:

M_{\max} = (14 \text{ kN})(2 \text{ m}) = 28 \text{ kN m}

Marking criteria: 1 mark for identifying that the shear step locates the point load at $x = 2$ m, 1 mark for the correct point-load magnitude (24 kN) as a check, 1 mark for correctly stating the maximum bending moment occurs where shear passes through zero, 1 mark for calculating the area under the shear diagram correctly, 1 mark for the correct final value (28 kN m) with units.

core4 marksThe pin support at the end of a truss provides an upward reaction of 18 kN. Only two members connect to this joint: a horizontal bottom-chord member (AB) and a diagonal member (AC) rising at 60 degrees above the horizontal. Using the method of joints (assume both members are in tension), calculate the force in each member and state whether each is actually in tension or compression.

Show worked solution →

Vertical equilibrium (the reaction acts up on the joint; the tension force in AC, assumed to pull away from the joint up the incline, also has an upward component):

18 + F_{AC}\sin 60^\circ = 0 \implies F_{AC} = \frac{-18}{\sin 60^\circ} = -20.8 \text{ kN}

The negative sign means AC is actually in compression (20.8 kN).

Horizontal equilibrium:

F_{AB} + F_{AC}\cos 60^\circ = 0 \implies F_{AB} = -(-20.8)(\cos 60^\circ) = 10.4 \text{ kN}

The positive sign confirms AB is in tension (10.4 kN), as assumed.

Marking criteria: 1 mark for a correctly signed vertical equilibrium equation including the external reaction, 1 mark for the correct magnitude of $F_{AC}$ (20.8 kN) identified as compression, 1 mark for the correct horizontal equilibrium equation, 1 mark for the correct magnitude of $F_{AB}$ (10.4 kN) identified as tension.

exam7 marksA pin-jointed footbridge truss spans between two supports and carries several downward panel-point loads on its top chord. Explain, with reference to equilibrium and buckling behaviour, why the top-chord members of such a truss are typically fabricated from a larger cross-section of steel than the bottom-chord members, and describe how a design engineer would determine the peak force each type of member must resist before selecting its size.

Show worked solution →

This is an extended-response question; markers reward a reasoned explanation linking equilibrium analysis to member design, not just a labelled diagram.

Why top and bottom chords carry opposite force types: In a simply supported truss loaded on its top chord, the overall structure behaves like a beam in bending: the top chord shortens (analogous to the compression side of a bent beam) while the bottom chord stretches (analogous to the tension side). The method of joints, applied panel by panel, confirms this: at each top joint the diagonal and chord forces resolve to give a negative (compressive) chord force, while at each bottom joint they resolve to a positive (tensile) chord force.
Why compression members need a larger cross-section: A tension member simply resists being pulled apart, so its required cross-sectional area is set only by the allowable tensile stress: $A = F / \sigma_{\text{allow}}$ . A compression member of the same force can fail well before reaching that stress by buckling, sudden sideways bowing that depends on the member's slenderness (length divided by a cross-sectional dimension), not just its area. To keep a compression member safely below its buckling load, the designer must increase its cross-section (or bracing) far beyond what tension alone would require, which is why top chords in trusses are visibly heavier sections than bottom chords carrying a similar magnitude of force.
Determining the peak force in each chord: The engineer first finds the support reactions from overall equilibrium ( $\sum F = 0$ , $\sum M = 0$ ), then applies the method of joints (or method of sections for a longer truss) working panel by panel from a joint with only two unknown members. The joint with the largest applied load and the smallest angle between chord and diagonal members typically produces the largest chord force; this governing panel sets the size for that chord type. Each chord is then sized to its own governing load case, tension chords against allowable tensile stress, compression chords against both allowable stress and the buckling load for their unbraced length.

Marker's note: top-band answers (1) explain the beam-like bending analogy to justify which chord is which, (2) explicitly name buckling, not just "compression is worse", as the mechanical reason for the larger section, (3) describe a concrete method (joints or sections) for finding the governing force, and (4) tie member sizing back to both stress and buckling criteria rather than stress alone.

exam6 marksA cantilever beam 4.0 m long is built into a wall at the left end. It carries a point load of 8 kN at the free end and a uniformly distributed load of 5 kN/m over the full length. Determine the vertical reaction and the fixing moment at the wall, and describe the shape of the bending moment diagram along the length of the beam.

Show worked solution →

UDL resultant: $W = wL = 5 \times 4.0 = 20$ kN, acting at the midspan, 2.0 m from the wall.

Vertical reaction (sum of vertical forces):

R = W + P = 20 + 8 = 28 \text{ kN (upward)}

Fixing moment (taking moments about the fixed wall):

M_{\text{wall}} = (20)(2.0) + (8)(4.0) = 40 + 32 = 72 \text{ kN m}

Shape of the bending-moment diagram. On a cantilever, the bending moment is zero at the free end and builds up towards the fixed end because more of the beam's own loading lies between the section and the free end as you move toward the wall. The point load alone would give a straight (linear) diagram; the UDL alone would give a curved (parabolic) diagram; combined, the diagram is a smoothly increasing curve from 0 kN m at the free end to the maximum of 72 kN m at the wall, with no change of sign anywhere along the span (a cantilever's bending moment never crosses zero along a single downward loading).

Marking criteria: 1 mark for the correct UDL resultant and location, 1 mark for the correct reaction (28 kN), 1 mark for a correctly set-out moment equation about the wall, 1 mark for the correct fixing moment (72 kN m) with units, 1 mark for identifying the maximum occurs at the wall, 1 mark for correctly describing the combined linear-plus-parabolic curved shape with no sign change.