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​ = 60 kNP2​ = 30 kNABRA​RB​ = ?1 m2 m1 m

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

L = 4 mP1 = 60 kN at 1 mP2 = 30 kN at 3 m
RB = kN
2.Beam support reactionsHard
w = 5 kN/mP = 50 kNABRA​ = ?RB​2 m2 m1 m

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

L = 5 mP = 50 kN at 4 mw = 5 kN/m from 2 to 4 m
RA = kN
3.Shear force & bending momentMedium
P1​ = 70 kNP2​ = 25 kNAB3 m1 m1 m

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

L = 5 mP1 = 70 kN at 3 mP2 = 25 kN at 4 m
Mmax = kN·m
4.Shear force & bending momentHard
w = 8 kN/mP = 40 kNABsection5 m3 m

Find the bending moment Mx at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 40 kN at 5 mw = 8 kN/m from 0 to 8 mx = 5 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB40 kN4 m1 m4 m

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

span = 4 mheight = 4 mB is 1 m from Avertical load = 40 kN
FAC = kN
6.Truss: method of jointsHard
ABCDEFGH25 kN20 kN20 kN30 kN35 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.

B: 25 kN ↓C: 20 kN ↓D: 20 kN ↓H: 30 kN ↓H: 35 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 = 10 kN/mP = 25 kNABδ1.5 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 = 25 kNw = 10 kN/mL = 1.5 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 8 kN/mδL = 8 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 = 8 kN/mL = 8 mE = 11 GPa (timber)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 6 m30 × 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 = 6 mE = 70 GPa (aluminium)b × h = 30 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 50 mmcross-section

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

L = 3 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RB = 37.5 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​ = 60 kNP2​ = 30 kNABRA​RB​1 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.

ΣMA = 0: RB·4 = 60·1 + 30·3
RB = 150/4 = 37.5 kN
ΣFy = 0: RA = 90 − 37.5 = 52.5 kN

Answer

RB = 37.5 kN
2.Beam support reactions

Answer: RA = 14 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 = 5 kN/mP = 50 kNABRA​RB​2 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 = 5·2 = 10 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 50·4 + 5·2·3
RB = 230/5 = 46 kN
ΣFy = 0: RA = 60 − 46 = 14 kN

Answer

RA = 14 kN
3.Shear force & bending moment

Answer: Mmax = 99 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·5 = 70·3 + 25·4
RB = 310/5 = 62 kN
ΣFy = 0: RA = 95 − 62 = 33 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·3 = 33·3 = 99 kN·m
under P2: M = RB·1 = 62·1 = 62 kN·m

Largest value

Mmax = 99 kN·m (under P1)

Answer

Mmax = 99 kN·m
4.Shear force & bending moment

Answer: Mx = 135 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.

UDL resultant: W = w·ℓ = 8·8 = 64 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 40·5 + 8·8·4
RB = 456/8 = 57 kN
ΣFy = 0: RA = 104 − 57 = 47 kN
w = 8 kN/mP = 40 kNABRA​ = 47RB​ = 57section5 m3 m

Cut at x = 5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 47·5 − 8·5·2.5 = 135 kN·m

Answer

Mx = 135 kN·m
5.Truss: method of joints

Answer: FAC = 7.5 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.

ABBCACACB40 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·4 = 40·1 → Cy = 10 kN
ΣFy = 0: Ay = 40 − 10 = 30 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 = 4/5 = 0.8, cos γ = 3/5 = 0.6
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −10/0.8 = −12.5 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−12.5)·0.6 = 7.5 kN

Answer

FAC = 7.5 kN (tension)
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: 4.687 kN (T), BC: 4.687 kN (T), CD: 31.56 kN (T), DE: 31.56 kN (T), FG: 60.63 kN (C), GH: 60.63 kN (C), AF: 66.15 kN (C), HE: 52.6 kN (C), BF: 25 kN (T), CG: 0, DH: 20 kN (T), FC: 34.9 kN (T), HC: 9.896 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δ = 3.076 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 = 25000 N, w = 10 N/mm, L = 1500 mm
EI = 1.12 × 10¹³ N·mm²

Each load on its own

δP = PL³/(3EI) = 25000·1500³/(3·1.12 × 10¹³) = 2.511 mm
δw = wL⁴/(8EI) = 10·1500⁴/(8·1.12 × 10¹³) = 0.565 mm

Add them

δ = 2.511 + 0.565 = 3.076 mm

Answer

δ = 3.076 mm
8.Beam deflection

Answer: Ireq = 1745 × 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 = 8000/360 = 22.22 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·8·8000⁴/(384·11000·22.22) = 1.745 × 10⁹ mm⁴
Ireq = 1745 × 10⁶ mm⁴

Answer

Ireq = 1745 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 2.591 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·135000/6000² = 2591 N = 2.591 kN

Answer

Pcr = 2.591 kN
10.Column buckling

Answer: σcr = 2.999 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/12.5 = 480

Critical stress

σcr = π²E/(KL/r)² = π²·70000/480² = 2.999 MPa

Answer

σcr = 2.999 MPa

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