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
P = 40 kNM = 20 kN·mABRB​ = ?5 m3 m2 m

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 8 m from A and a clockwise couple M at 5 m from A. Find the vertical reaction RB, taking upward as positive.

L = 10 mP = 40 kN at 8 mM = 20 kN·m (clockwise) at 5 m
RB = kN
2.Beam support reactionsHard
P = 25 kNQ = 80 kNADCRD​ = ?1 m2 m1 m4 m

A compound beam is fixed to the wall at A, has an internal hinge at C (3 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 4 m from A and a point load Q at 1 m from A. Find the reaction RD at the roller.

AC = 3 mAD = 8 mP = 25 kN at 4 mQ = 80 kN at 1 m
RD = kN
3.Shear force & bending momentMedium
25 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.

25 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
w = 16 kN/mAB5 m2 m1 m

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.

L = 8 mw = 16 kN/m from 5 to 7 m
Mmax = kN·m
5.Truss: method of jointsMedium
ABCDEFGH20 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.

D: 20 kN ↓G: 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
ABCDEFGH15 kN35 kN25 kN25 kN20 kN40 kN4 × 3 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: 15 kN ↓C: 35 kN ↓D: 25 kN ↓F: 25 kN ↓F: 20 kN →H: 40 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
7.Beam deflectionMedium
w = 20 kN/mP = 9 kNABδ2 m

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 9 kNw = 20 kN/mL = 2 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 10 kN/mP = 24 kNδL = 2.5 m

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δmax.

P = 24 kNw = 10 kN/mL = 2.5 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 2 m50 × 80 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 25 kNL = 1 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 25 kNn = 2L = 1 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 34 kN

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.

Moments about A

Counter-clockwise positive.

ΣMA = 0: RB·10 − 40·8 − 20 = 0
RB = 340/10 = 34 kN

Vertical equilibrium

RA = 40 − 34 = 6 kN

Answer

RB = 34 kN

A negative reaction acts downward.

2.Beam support reactions

Answer: RD = 5 kN

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

ΣMC = 0: RD·5 = 25·1
RD = 5 kN
VC = 25 − 5 = 20 kN

Answer

RD = 5 kN
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 = 24.17 kN, R2 = 60.83 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 = 48.33, MC = 96.67, MD = 145, ME = 193.3 kN·m, MF = 121.7, MG = 0

Answer

Mark E: ME = 193.3 kN·m
4.Shear force & bending moment

Answer: Mmax = 42 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 16·2 = 32 kN, acting at x̄ = 6 m
ΣMA = 0: RB·8 = 16·2·6
RB = 192/8 = 24 kN
ΣFy = 0: RA = 32 − 24 = 8 kN

Zero shear

The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.

RA − w·(x0 − 5) = 0
x0 = 5 + 8/16 = 5.5 m

Maximum moment

Mmax = RA·x0 − w·(x0 − 5)²/2 = 8·5.5 − 16·0.5²/2 = 42 kN·m

Answer

Mmax = 42 kN·m
5.Truss: method of joints

Answer: member BF

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 B

AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.

ΣF⊥ = 0 → FBF = 0

Check with the method of joints

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

AB: 20 kN (T), BC: 20 kN (T), CD: 33.33 kN (T), DE: 33.33 kN (T), FG: 40 kN (C), GH: 40 kN (C), AF: 25 kN (C), HE: 41.67 kN (C), BF: 0, CG: 20 kN (C), DH: 20 kN (T), FC: 25 kN (T), HC: 8.333 kN (T)

Answer

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

Answer: member CG

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 G

FG and GH lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.

ΣF⊥ = 0 → FCG = 0

Check with the method of joints

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

AB: 78.75 kN (T), BC: 78.75 kN (T), CD: 81.25 kN (T), DE: 81.25 kN (T), FG: 97.5 kN (C), GH: 97.5 kN (C), AF: 83.09 kN (C), HE: 114.9 kN (C), BF: 15 kN (T), CG: 0, DH: 25 kN (T), FC: 26.52 kN (T), HC: 22.98 kN (T)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δ = 5.614 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.

P = 9000 N, w = 20 N/mm, L = 2000 mm
EI = 1.14 × 10¹³ N·mm²

Each load on its own

δP = PL³/(3EI) = 9000·2000³/(3·1.14 × 10¹³) = 2.105 mm
δw = wL⁴/(8EI) = 20·2000⁴/(8·1.14 × 10¹³) = 3.509 mm

Add them

δ = 2.105 + 3.509 = 5.614 mm

Answer

δ = 5.614 mm
8.Beam deflection

Answer: δmax = 1.791 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.

P = 24000 N, w = 10 N/mm, L = 2500 mm, EI = 7.2 × 10¹² N·mm²

Each load on its own

δP = PL³/(48EI) = 24000·2500³/(48·7.2 × 10¹²) = 1.085 mm
δw = 5wL⁴/(384EI) = 5·10·2500⁴/(384·7.2 × 10¹²) = 0.7064 mm

Add them

δmax = 1.085 + 0.7064 = 1.791 mm

Answer

δmax = 1.791 mm
9.Column buckling

Answer: Pcr = 293.7 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 = 80·50³/12 = 833333 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·833333/1400² = 293738 N = 293.7 kN

Answer

Pcr = 293.7 kN
10.Column buckling

Answer: d = 49.28 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 = 2·25000 = 50000 N
KL = 2·1000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·2000²/(π³·70000))^(1/4) = 49.28 mm

Answer

d = 49.28 mm

Check: the slenderness KL/r is 162.3, so the column is slender and Euler applies.

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