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​ = 70 kNP2​ = 15 kNABCRB​ = ?3 m3 m4 m

The beam has a pin at A and a roller at B, 6 m from A, and overhangs B to the free end C. Find the vertical reaction RB, taking upward as positive.

AB = 6 mAC = 10 mP1 = 70 kN at 3 mP2 = 15 kN at 10 m
RB = kN
2.Beam support reactionsHard
P = 55 kN, 40°Q = 50 kNABRA​ = ?1 m1.5 m1.5 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 1 m from A. Find the magnitude of the total reaction force RA at the pin A.

L = 4 mP = 55 kN at 2.5 m, θ = 40°Q = 50 kN at 1 m
RA = kN
3.Shear force & bending momentMedium
P1​ = 75 kNP2​ = 40 kNABC3 m3 m1.5 m

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment Mx under the load P1. Use sagging moments positive and upward shear on the left face positive.

L = 7.5 mP1 = 75 kN at 3 mP2 = 40 kN at 7.5 msupports at x = 0 and x = 6 m
Mx = kN·m
4.Shear force & bending momentHard
w = 10 kN/mP = 60 kNAB2 m6 m

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.

L = 8 mP = 60 kN at 2 mw = 10 kN/m from 0 to 8 m
Mmax = kN·m
5.Truss: method of jointsMedium
ABBCACACB25 kN5 m2 m2 m

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

span = 5 mheight = 2 mB is 2 m from Avertical load = 25 kN
FAC = kN
6.Truss: method of jointsHard
ABBCACACB60 kN10 kN4 m3 m4 m

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

span = 4 mheight = 4 mB is 3 m from Avertical load = 60 kNhorizontal load = 10 kN ←
FAB = kN
7.Beam deflectionMedium
w = 5 kN/mδL = 3 m

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

w = 5 kN/mL = 3 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 12 kN/mP = 18 kNδL = 2 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 = 18 kNw = 12 kN/mL = 2 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 2 md = 60 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 110 kNL = 6 md = ?cross-section

A solid round column of length L made of aluminium is fixed at both ends. It must carry 110 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 110 kNn = 2L = 6 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 60 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·6 = 70·3 + 15·10
RB = 360/6 = 60 kN

Vertical equilibrium

RA = 70 + 15 − 60 = 25 kN

Answer

RB = 60 kN
2.Beam support reactions

Answer: RA = 65.97 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.

Components of P

Pv = 55·sin 40° = 35.35 kN
Ph = 55·cos 40° = 42.13 kN

Reactions

ΣMA = 0: RB·4 = 35.35·2.5 + 50·1 → RB = 34.6 kN
Ay = 35.35 + 50 − 34.6 = 50.76 kN
Ax = Ph = 42.13 kN

Resultant at A

RA = √(42.13² + 50.76²) = 65.97 kN

Answer

RA = 65.97 kN
3.Shear force & bending moment

Answer: Mx = 82.5 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.

Support reactions

Equilibrium of the whole beam. The overhang load also acts about A.

ΣMA = 0: RB·6 = 75·3 + 40·7.5
RB = 525/6 = 87.5 kN
ΣFy = 0: RA = 115 − 87.5 = 27.5 kN

Cut under P1 and keep the left part

Mx = RA·3 = 27.5·3 = 82.5 kN·m

Answer

Mx = 82.5 kN·m
4.Shear force & bending moment

Answer: Mmax = 151.3 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 = 10·8 = 80 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 60·2 + 10·8·4
RB = 440/8 = 55 kN
ΣFy = 0: RA = 140 − 55 = 85 kN

Where is V = 0?

RA − P − w·x0 = 0 → x0 = (85 − 60)/10 = 2.5 m

Maximum moment

Mmax = RA·x0 − P·(x0 − 2) − w·x0²/2 = 151.3 kN·m

Answer

Mmax = 151.3 kN·m

It occurs at x = 2.5 m from A.

5.Truss: method of joints

Answer: FAC = 15 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.

ABBCACACB25 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·5 = 25·2 → Cy = 10 kN
ΣFy = 0: Ay = 25 − 10 = 15 kN

Joint C

The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.

sin γ = 2/3.606 = 0.5547, cos γ = 3/3.606 = 0.8321
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −10/0.5547 = −18.03 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−18.03)·0.8321 = 15 kN

Answer

FAC = 15 kN (tension)
6.Truss: method of joints

Answer: FAB = −31.25 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.

ABBCACACB60 kN10 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 60·3 − 10·4 → Cy = 35 kN
ΣFy = 0: Ay = 60 − 35 = 25 kN
ΣFx = 0: Ax = 10 kN

Joint A

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

sin α = 4/5 = 0.8
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −25/0.8 = −31.25 kN

Answer

FAB = −31.25 kN (compression)
7.Beam deflection

Answer: δ = 21.09 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 = 5 N/mm, L = 3000 mm
EI = 2.4 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 5·3000⁴/(8·2.4 × 10¹²) = 21.09 mm

Answer

δ = 21.09 mm
8.Beam deflection

Answer: δmax = 13.89 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 = 18000 N, w = 12 N/mm, L = 2000 mm, EI = 3.96 × 10¹¹ N·mm²

Each load on its own

δP = PL³/(48EI) = 18000·2000³/(48·3.96 × 10¹¹) = 7.576 mm
δw = 5wL⁴/(384EI) = 5·12·2000⁴/(384·3.96 × 10¹¹) = 6.313 mm

Add them

δmax = 7.576 + 6.313 = 13.89 mm

Answer

δmax = 13.89 mm
9.Column buckling

Answer: Pcr = 109.9 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/2000² = 109878 N = 109.9 kN

Answer

Pcr = 109.9 kN
10.Column buckling

Answer: d = 87.41 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 = 2·110000 = 220000 N
KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·220000·3000²/(π³·70000))^(1/4) = 87.41 mm

Answer

d = 87.41 mm

Check: the slenderness KL/r is 137.3, so the column is slender and Euler applies.

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