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​ = 55 kNP2​ = 10 kNABRA​RB​ = ?1 m1 m2 m

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

L = 4 mP1 = 55 kN at 1 mP2 = 10 kN at 2 m
RB = kN
2.Beam support reactionsHard
w = 5 kN/mP = 40 kNQ = 75 kNADCRD​ = ?1 m2 m2 m5 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 (10 m from A). It carries a point load P at 5 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 1 m from A. Find the reaction RD at the roller.

AC = 3 mAD = 10 mP = 40 kN at 5 mw = 5 kN/m from 3 to 10 mQ = 75 kN at 1 m
RD = kN
3.Shear force & bending momentMedium
10 kN60 kNABCDEFG1 m1 m1 m1 m1 m1 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.

10 kN @ 2 m60 kN @ 4 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
4.Shear force & bending momentHard
12 kN/m10 kN30 kN35 kNABCDEFG1 m1 m1 m1 m1 m1 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.

10 kN @ 0 m30 kN @ 6 m35 kN @ 2 m12 kN/m, 1–4 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
5.Truss: method of jointsMedium
ABCDEFGH10 kN15 kN20 kN10 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: 10 kN ↓F: 15 kN ↓G: 20 kN ↓H: 10 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 kN10 kN15 kN30 kN20 kN4 × 3 m4 m

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

C: 15 kN ↓D: 10 kN ↓F: 15 kN ↓F: 30 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
7.Beam deflectionMedium
P = 24 kNABδ1.5 m1 m

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1.5 m from A. Find the deflection δB at the free end B.

P = 24 kNa = 1.5 mL = 2.5 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
δB = mm
8.Beam deflectionHard
w = 2 kN/mP = 19 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 = 19 kNw = 2 kN/mL = 2.5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 2 md = 50 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)d = 50 mm
Pcr = kN
10.Column bucklingHard
P = 120 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 120 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 120 kNn = 2L = 3 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 18.75 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​ = 55 kNP2​ = 10 kNABRA​RB​1 m1 m2 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 = 55·1 + 10·2
RB = 75/4 = 18.75 kN
ΣFy = 0: RA = 65 − 18.75 = 46.25 kN

Answer

RB = 18.75 kN
2.Beam support reactions

Answer: RD = 28.93 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.

W = 5·7 = 35 kN at 3.5 m from C
ΣMC = 0: RD·7 = 40·2 + 35·3.5
RD = 28.93 kN
VC = 75 − 28.93 = 46.07 kN

Answer

RD = 28.93 kN
3.Shear force & bending moment

Answer: E (x = 4 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 = 26.67 kN, R2 = 43.33 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 = 26.67, MC = 53.33, MD = 70, ME = 86.67 kN·m, MF = 43.33, MG = 0

Answer

Mark E: ME = 86.67 kN·m
4.Shear force & bending moment

Answer: E (x = 4 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 = 34.67 kN, R2 = 76.33 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 = −10, MC = 8.667, MD = −19.67, ME = −60 kN·m, MF = −30, MG = 0

Answer

Mark E: ME = −60 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: 41.67 kN (T), BC: 41.67 kN (T), CD: 31.67 kN (T), DE: 31.67 kN (T), FG: 50 kN (C), GH: 50 kN (C), AF: 52.08 kN (C), HE: 39.58 kN (C), BF: 10 kN (T), CG: 20 kN (C), DH: 0, FC: 10.42 kN (T), HC: 22.92 kN (T)

Answer

Zero-force member: DH
6.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: 45.94 kN (T), BC: 45.94 kN (T), CD: 29.06 kN (T), DE: 29.06 kN (T), FG: 50.63 kN (C), GH: 50.63 kN (C), AF: 26.56 kN (C), HE: 48.44 kN (C), BF: 0, CG: 20 kN (C), DH: 10 kN (T), FC: 7.813 kN (T), HC: 35.94 kN (T)

Answer

Zero-force member: BF
7.Beam deflection

Answer: δB = 6.429 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, a = 1500 mm, L = 2500 mm
EI = 8.4 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δP = Pa³/(3EI) = 3.214 mm
θP = Pa²/(2EI) = 0.003214 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 6.429 mm

Answer

δB = 6.429 mm
8.Beam deflection

Answer: δmax = 11.49 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 = 19000 N, w = 2 N/mm, L = 2500 mm, EI = 6.27 × 10¹¹ N·mm²

Each load on its own

δP = PL³/(48EI) = 19000·2500³/(48·6.27 × 10¹¹) = 9.864 mm
δw = 5wL⁴/(384EI) = 5·2·2500⁴/(384·6.27 × 10¹¹) = 1.622 mm

Add them

δmax = 9.864 + 1.622 = 11.49 mm

Answer

δmax = 11.49 mm
9.Column buckling

Answer: Pcr = 108.1 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 round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·306796/1400² = 108141 N = 108.1 kN

Answer

Pcr = 108.1 kN
10.Column buckling

Answer: d = 68.71 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 = 2·120000 = 240000 N
KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·240000·3000²/(π³·200000))^(1/4) = 68.71 mm

Answer

d = 68.71 mm

Check: the slenderness KL/r is 174.6, so the column is slender and Euler applies.

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