Structural Analysis: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Beam support reactions

Equilibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Uniform load
resultant w·L at the middle of the load
Moment
M = F·d

Method

  1. Draw a free-body diagram and replace each support by its reactions.
  2. Replace every distributed load by a single force at its centroid.
  3. Take moments about one support to find the reaction at the other.
  4. Use ΣFy = 0 to find the remaining reaction, and check with a second moment equation.

Shear force & bending moment

Shear and moment
V = ΣF on one side, M = ΣF·d on one side
Bending stress
σ = M·y/I
Rectangle
I = bh³/12
Section modulus
W = I/ymax
Simply supported, uniform load
Mmax = wL²/8

Method

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.

Truss: method of joints

Method of joints
ΣFx = 0, ΣFy = 0 at each joint
Method of sections
ΣM = 0 about a cut point
Member force
positive = tension, negative = compression

Method

  1. Find the support reactions for the whole truss.
  2. Method of joints: start at a joint with at most two unknown member forces and use ΣFx = 0 and ΣFy = 0.
  3. Method of sections: cut through the member you need and take moments about a point where the other unknown members meet.
  4. Assume tension; a negative answer means compression.

Beam deflection

Point load at midspan
δ = PL³/(48EI)
Uniform load, simply supported
δ = 5wL⁴/(384EI)
Cantilever, end load
δ = PL³/(3EI)
Cantilever, uniform load
δ = wL⁴/(8EI)
Superposition
δ = δ1 + δ2

Column buckling

Euler load
Pcr = π²EI/(KL)²
Effective length factor
K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress
σcr = Pcr/A
Slenderness
λ = KL/r, r = √(I/A)

2Practice problems

1.Beam support reactionsMedium
P1​ = 10 kNP2​ = 55 kNABRA​RB​ = ?1.5 m1 m1.5 m

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.

L = 4 mP1 = 10 kN at 1.5 mP2 = 55 kN at 2.5 m
RB = kN
2.Beam support reactionsHard
w0​ = 12 kN/mP = 40 kNABMA​ = ?1.5 m2.5 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 12 kN/m at the wall and falls to zero at the free end B, plus a point load P at 1.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 4 mw0 = 12 kN/mP = 40 kN at 1.5 m
MA = kN·m
3.Shear force & bending momentMedium
35 kN60 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

35 kN @ 10 m60 kN @ 8 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
4.Shear force & bending momentHard
10 kN/m55 kN15 kN40 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

55 kN @ 0 m15 kN @ 12 m40 kN @ 4 m10 kN/m, 2–8 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
5.Truss: method of jointsMedium
ABCDEFGH20 kN20 kN15 kN20 kN4 × 4 m3 m

Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.

B: 20 kN ↓C: 20 kN ↓G: 15 kN ↓H: 20 kN ↓
  1. 1member AB
  2. 2member BC
  3. 3member CD
  4. 4member DE
  5. 5member FG
  6. 6member GH
  7. 7member AF
  8. 8member HE
  9. 9member BF
  10. 10member CG
  11. 11member DH
  12. 12member FC
  13. 13member HC
6.Truss: method of jointsHard
ABBCACACB35 kN20 kN5 m1 m4 m

Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.

span = 5 mheight = 4 mB is 1 m from Avertical load = 35 kNhorizontal load = 20 kN →
FAB = kN
7.Beam deflectionMedium
w = 4 kN/mδL = 4 m

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 4 kN/mL = 4 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
P = 50 kNAB3 m3 m

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction RB at the prop.

L = 6 mP = 50 kN at 3 m
RB = kN
9.Column bucklingMedium
PL = 5 m40 × 60 mmcross-section

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 5 mE = 70 GPa (aluminium)b × h = 40 × 60 mm
Pcr = kN
10.Column bucklingHard
P = 80 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is fixed at both ends. It must carry 80 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 80 kNn = 2L = 2.5 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 38.13 kN

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.

P1​ = 10 kNP2​ = 55 kNABRA​RB​1.5 m1 m1.5 m

Free-body diagram

Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·4 = 10·1.5 + 55·2.5
RB = 152.5/4 = 38.13 kN
ΣFy = 0: RA = 65 − 38.13 = 26.88 kN

Answer

RB = 38.13 kN
2.Beam support reactions

Answer: MA = 92 kN·m

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.

Resultant of the triangular load

W = ½·12·4 = 24 kN
x̄ = L/3 = 1.333 m

Moment about the wall

MA = W·x̄ + P·a = 24·1.333 + 40·1.5 = 92 kN·m

Answer

MA = 92 kN·m
3.Shear force & bending moment

Answer: E (x = 8 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 25.83 kN, R2 = 69.17 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 51.67, MC = 103.3, MD = 155, ME = 206.7 kN·m, MF = 138.3, MG = 0

Answer

Mark E: ME = 206.7 kN·m
4.Shear force & bending moment

Answer: B (x = 2 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 120 kN, R2 = 50 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = −110 kN·m, MC = 0, MD = −10, ME = −60, MF = −30, MG = 0

Answer

Mark B: MB = −110 kN·m
5.Truss: method of joints

Answer: member DH

Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.

Joint D

CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.

ΣF⊥ = 0 → FDH = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 50 kN (T), BC: 50 kN (T), CD: 50 kN (T), DE: 50 kN (T), FG: 73.33 kN (C), GH: 73.33 kN (C), AF: 62.5 kN (C), HE: 62.5 kN (C), BF: 20 kN (T), CG: 15 kN (C), DH: 0, FC: 29.17 kN (T), HC: 29.17 kN (T)

Answer

Zero-force member: DH
6.Truss: method of joints

Answer: FAB = −12.37 kN

Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.

ABBCACACB35 kN20 kNAy​Cy​Ax​

Support reactions

Equilibrium of the whole truss. Moments about A, counter-clockwise positive.

ΣMA = 0: Cy·5 = 35·1 + 20·4 → Cy = 23 kN
ΣFy = 0: Ay = 35 − 23 = 12 kN
ΣFx = 0: Ax = −20 kN

Joint A

Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.

sin α = 4/4.123 = 0.9701
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −12/0.9701 = −12.37 kN

Answer

FAB = −12.37 kN (compression)
7.Beam deflection

Answer: δ = 79.5 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 4 N/mm, L = 4000 mm
EI = 1.61 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 4·4000⁴/(8·1.61 × 10¹²) = 79.5 mm

Answer

δ = 79.5 mm
8.Beam deflection

Answer: RB = 15.63 kN

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition). For a statically indeterminate beam, remove a redundant support and require the deflection there to be zero (compatibility).

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δload = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force RB alone lifts the free end by RBL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

RB·L³/(3EI) = 5PL³/(48EI) → RB = 5P/16 = 5·50/16 = 15.63 kN

Answer

RB = 15.63 kN
9.Column buckling

Answer: Pcr = 2.211 kN

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

Imin = h·b³/12 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 2·5000 = 10000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·320000/10000² = 2211 N = 2.211 kN

Answer

Pcr = 2.211 kN
10.Column buckling

Answer: d = 52.11 mm

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Required critical load

Pcr = n·P = 2·80000 = 160000 N
KL = 0.5·2500 = 1250 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·160000·1250²/(π³·70000))^(1/4) = 52.11 mm

Answer

d = 52.11 mm

Check: the slenderness KL/r is 95.96, so the column is slender and Euler applies.

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