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 = 50 kNM = 30 kN·mABRA​ = ?0.5 m4 m0.5 m

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 4.5 m from A and a counter-clockwise couple M at 0.5 m from A. Find the vertical reaction RA, taking upward as positive.

L = 5 mP = 50 kN at 4.5 mM = 30 kN·m (counter-clockwise) at 0.5 m
RA = kN
2.Beam support reactionsHard
w = 4 kN/mP = 50 kNABRA​RB​ = ?1 m1 m2 m1 m

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

L = 5 mP = 50 kN at 1 mw = 4 kN/m from 2 to 4 m
RB = kN
3.Shear force & bending momentMedium
P1​ = 20 kNP2​ = 40 kNAB6 m1 m3 m

The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.

L = 10 mP1 = 20 kN at 6 mP2 = 40 kN at 7 m
Mmax = kN·m
4.Shear force & bending momentHard
30 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.

30 kN @ 12 m60 kN @ 6 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
ABCDEFGH15 kN40 kN40 kN10 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 ↓D: 40 kN ↓F: 40 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 BG
  13. 13member DG
6.Truss: method of jointsHard
ABCDEFGH25 kN10 kN30 kN30 kN30 kN30 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: 25 kN ↓D: 10 kN ↓F: 30 kN ↓F: 30 kN →G: 30 kN ↓H: 30 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 m5 m

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δB at the free end B.

P = 24 kNa = 1 mL = 6 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴
δB = mm
8.Beam deflectionHard
w = 2.5 kN/mABδ6 m

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth hreq of the section.

w = 2.5 kN/mL = 6 mE = 11 GPa (timber)b = 75 mm
hreq = mm
9.Column bucklingMedium
PL = 5 md = 40 mmcross-section

A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 5 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
P = 130 kNL = 2 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 130 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 130 kNn = 2L = 2 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 11 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·5 − 50·4.5 + 30 = 0
RB = 195/5 = 39 kN

Vertical equilibrium

RA = 50 − 39 = 11 kN

Answer

RA = 11 kN

A negative reaction acts downward.

2.Beam support reactions

Answer: RB = 14.8 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.

w = 4 kN/mP = 50 kNABRA​RB​1 m1 m2 m1 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.

UDL resultant: W = w·ℓ = 4·2 = 8 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 50·1 + 4·2·3
RB = 74/5 = 14.8 kN
ΣFy = 0: RA = 58 − 14.8 = 43.2 kN

Answer

RB = 14.8 kN
3.Shear force & bending moment

Answer: Mmax = 120 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.

ΣMA = 0: RB·10 = 20·6 + 40·7
RB = 400/10 = 40 kN
ΣFy = 0: RA = 60 − 40 = 20 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·6 = 20·6 = 120 kN·m
under P2: M = RB·3 = 40·3 = 120 kN·m

Largest value

Mmax = 120 kN·m (under P1)

Answer

Mmax = 120 kN·m
4.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 = 0 kN, R2 = 90 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 = 0, MC = 0, MD = 0, ME = −120 kN·m, MF = −60, MG = 0

Answer

Mark E: ME = −120 kN·m
5.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 C

BC and CD 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: 53.75 kN (T), BC: 52.5 kN (T), CD: 52.5 kN (T), DE: 51.25 kN (T), FG: 53.75 kN (C), GH: 51.25 kN (C), AF: 76.01 kN (C), HE: 72.48 kN (C), BF: 13.75 kN (T), CG: 0, DH: 41.25 kN (T), BG: 1.768 kN (T), DG: 1.768 kN (C)

Answer

Zero-force member: CG
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: 67.5 kN (T), BC: 67.5 kN (T), CD: 56.25 kN (T), DE: 56.25 kN (T), FG: 82.5 kN (C), GH: 82.5 kN (C), AF: 62.5 kN (C), HE: 93.75 kN (C), BF: 0, CG: 30 kN (C), DH: 10 kN (T), FC: 25 kN (T), HC: 43.75 kN (T)

Answer

Zero-force member: BF
7.Beam deflection

Answer: δB = 26.98 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 = 1000 mm, L = 6000 mm
EI = 2.52 × 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) = 3.175 mm
θP = Pa²/(2EI) = 0.004762 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 26.98 mm

Answer

δB = 26.98 mm
8.Beam deflection

Answer: hreq = 294.6 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 = 2.5 N/mm, L = 6000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/250 = 6000/250 = 24 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·2.5·6000⁴/(384·11000·24) = 1.598 × 10⁸ mm⁴

Required depth

bh³/12 = Ireq → h = (12·1.598 × 10⁸/75)1/3 = 294.6 mm

Answer

hreq = 294.6 mm

In practice, choose the next standard depth above this value.

9.Column buckling

Answer: Pcr = 39.69 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/2500² = 39688 N = 39.69 kN

Answer

Pcr = 39.69 kN
10.Column buckling

Answer: d = 80.95 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 = 2·130000 = 260000 N
KL = 2·2000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·260000·4000²/(π³·200000))^(1/4) = 80.95 mm

Answer

d = 80.95 mm

Check: the slenderness KL/r is 197.7, so the column is slender and Euler applies.

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