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​ = 20 kNP2​ = 55 kNABRA​RB​ = ?1 m3 m1 m

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

L = 5 mP1 = 20 kN at 1 mP2 = 55 kN at 4 m
RB = kN
2.Beam support reactionsHard
w = 3 kN/mP = 10 kNABRA​RB​ = ?2 m1 m1 m

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

L = 4 mP = 10 kN at 2 mw = 3 kN/m from 0 to 3 m
RB = kN
3.Shear force & bending momentMedium
P1​ = 40 kNP2​ = 45 kNAB1 m2 m5 m

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

L = 8 mP1 = 40 kN at 1 mP2 = 45 kN at 3 m
Mmax = kN·m
4.Shear force & bending momentHard
w = 24 kN/mAB1 m4 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 = 5 mw = 24 kN/m from 1 to 5 m
Mmax = kN·m
5.Truss: method of jointsMedium
ABBCACACB25 kN8 m7 m4 m

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

span = 8 mheight = 4 mB is 7 m from Avertical load = 25 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB20 kN20 kN4 m3 m2 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 = 2 mB is 3 m from Avertical load = 20 kNhorizontal load = 20 kN →
FAB = kN
7.Beam deflectionMedium
w = 5 kN/mδL = 3 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.

w = 5 kN/mL = 3 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 4 kN/mδL = 3 m

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.

w = 4 kN/mL = 3 mE = 200 GPa (steel)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 1 m30 × 40 mmcross-section

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

L = 1 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 80 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 5 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RB = 48 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​ = 20 kNP2​ = 55 kNABRA​RB​1 m3 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.

ΣMA = 0: RB·5 = 20·1 + 55·4
RB = 240/5 = 48 kN
ΣFy = 0: RA = 75 − 48 = 27 kN

Answer

RB = 48 kN
2.Beam support reactions

Answer: RB = 8.375 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 = 3 kN/mP = 10 kNABRA​RB​2 m1 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 = 3·3 = 9 kN, acting at x̄ = 1.5 m
ΣMA = 0: RB·4 = 10·2 + 3·3·1.5
RB = 33.5/4 = 8.375 kN
ΣFy = 0: RA = 19 − 8.375 = 10.63 kN

Answer

RB = 8.375 kN
3.Shear force & bending moment

Answer: Mmax = 109.4 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·8 = 40·1 + 45·3
RB = 175/8 = 21.88 kN
ΣFy = 0: RA = 85 − 21.88 = 63.13 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·1 = 63.13·1 = 63.13 kN·m
under P2: M = RB·5 = 21.88·5 = 109.4 kN·m

Largest value

Mmax = 109.4 kN·m (under P2)

Answer

Mmax = 109.4 kN·m
4.Shear force & bending moment

Answer: Mmax = 69.12 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 = 24·4 = 96 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 24·4·3
RB = 288/5 = 57.6 kN
ΣFy = 0: RA = 96 − 57.6 = 38.4 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 − 1) = 0
x0 = 1 + 38.4/24 = 2.6 m

Maximum moment

Mmax = RA·x0 − w·(x0 − 1)²/2 = 38.4·2.6 − 24·1.6²/2 = 69.12 kN·m

Answer

Mmax = 69.12 kN·m
5.Truss: method of joints

Answer: FAB = −6.299 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·8 = 25·7 → Cy = 21.88 kN
ΣFy = 0: Ay = 25 − 21.88 = 3.125 kN

Joint A

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

sin α = 4/8.062 = 0.4961
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −3.125/0.4961 = −6.299 kN

Answer

FAB = −6.299 kN (compression)
6.Truss: method of joints

Answer: FAB = 9.014 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.

ABBCACACB20 kN20 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 20·3 + 20·2 → Cy = 25 kN
ΣFy = 0: Ay = 20 − 25 = −5 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 = 2/3.606 = 0.5547
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −−5/0.5547 = 9.014 kN

Answer

FAB = 9.014 kN (tension)
7.Beam deflection

Answer: δmax = 13.32 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 = 3.96 × 10¹¹ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·5·3000⁴/(384·3.96 × 10¹¹) = 13.32 mm

Answer

δmax = 13.32 mm
8.Beam deflection

Answer: Ireq = 2.531 × 10⁶ mm⁴

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).

Deflection limit

δallow = L/360 = 3000/360 = 8.333 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·4·3000⁴/(384·200000·8.333) = 2.531 × 10⁶ mm⁴
Ireq = 2.531 × 10⁶ mm⁴

Answer

Ireq = 2.531 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 44.41 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/2000² = 44413 N = 44.41 kN

Answer

Pcr = 44.41 kN
10.Column buckling

Answer: σcr = 22.56 MPa

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·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 0.7·5000 = 3500 mm
KL/r = 3500/20 = 175

Critical stress

σcr = π²E/(KL/r)² = π²·70000/175² = 22.56 MPa

Answer

σcr = 22.56 MPa

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