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
w0​ = 4 kN/mP = 10 kNABRB​ = ?2.5 m2.5 m

The simply supported beam (pin at A, roller at B) carries a linearly varying load that falls from 4 kN/m at A to zero at B. A point load P also acts at 2.5 m from A. Find the vertical support reaction RB.

L = 5 mw0 = 4 kN/mP = 10 kN at 2.5 m
RB = kN
2.Beam support reactionsHard
P = 55 kN, 45°Q = 45 kNABRA​ = ?3.5 m2 m0.5 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 5.5 m from A, inclined at 45° to the horizontal, and a vertical load Q acts at 3.5 m from A. Find the magnitude of the total reaction force RA at the pin A.

L = 6 mP = 55 kN at 5.5 m, θ = 45°Q = 45 kN at 3.5 m
RA = kN
3.Shear force & bending momentMedium
w = 10 kN/mP = 20 kNABsection0.5 m2.5 m

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 3 m from A. Find the bending moment Mx at the section x = 0.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 3 mP = 20 kN at 3 mw = 10 kN/m from 0 to 3 mx = 0.5 m
Mx = kN·m
4.Shear force & bending momentHard
w = 8 kN/mP = 50 kNABsection1.5 m1.5 m3 m

Find the bending moment Mx at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 50 kN at 1.5 mw = 8 kN/m from 0 to 6 mx = 3 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB30 kN8 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 = 8 mheight = 2 mB is 3 m from Avertical load = 30 kN
FAB = kN
6.Truss: method of jointsHard
ABCDEFGH20 kN40 kN40 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: 20 kN ↓C: 40 kN ↓D: 40 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 = 4 kN/mδL = 8 m

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

w = 4 kN/mL = 8 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 15 kN/mP = 30 kNδL = 1.5 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 = 30 kNw = 15 kN/mL = 1.5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 2 md = 30 mmcross-section

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

L = 2 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 40 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 40 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RB = 8.333 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.

Resultant of the triangular load

W = ½·4·5 = 10 kN
x̄ = L/3 = 1.667 m

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·5 = 10·1.667 + 10·2.5
RB = 8.333 kN
RA = 20 − 8.333 = 11.67 kN

Answer

RB = 8.333 kN
2.Beam support reactions

Answer: RA = 44.68 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 45° = 38.89 kN
Ph = 55·cos 45° = 38.89 kN

Reactions

ΣMA = 0: RB·6 = 38.89·5.5 + 45·3.5 → RB = 61.9 kN
Ay = 38.89 + 45 − 61.9 = 21.99 kN
Ax = Ph = 38.89 kN

Resultant at A

RA = √(38.89² + 21.99²) = 44.68 kN

Answer

RA = 44.68 kN
3.Shear force & bending moment

Answer: Mx = −81.25 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.

Keep the free-end part

P: −20·2.5 = −50 kN·m
w: −10·2.5²/2 = −31.25 kN·m
Mx = −50 − 31.25 = −81.25 kN·m

Answer

Mx = −81.25 kN·m

Negative means hogging: tension on the top face.

4.Shear force & bending moment

Answer: Mx = 73.5 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.

UDL resultant: W = w·ℓ = 8·6 = 48 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 50·1.5 + 8·6·3
RB = 219/6 = 36.5 kN
ΣFy = 0: RA = 98 − 36.5 = 61.5 kN
w = 8 kN/mP = 50 kNABRA​ = 61.5RB​ = 36.5section1.5 m1.5 m3 m

Cut at x = 3 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 = 61.5·3 − 50·(3 − 1.5) − 8·3·1.5 = 73.5 kN·m

Answer

Mx = 73.5 kN·m
5.Truss: method of joints

Answer: FAB = −33.8 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.

ABBCACACB30 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·8 = 30·3 → Cy = 11.25 kN
ΣFy = 0: Ay = 30 − 11.25 = 18.75 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 = −18.75/0.5547 = −33.8 kN

Answer

FAB = −33.8 kN (compression)
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: 7.5 kN (T), BC: 7.5 kN (T), CD: 32.5 kN (T), DE: 32.5 kN (T), FG: 70 kN (C), GH: 70 kN (C), AF: 70.83 kN (C), HE: 54.17 kN (C), BF: 20 kN (T), CG: 0, DH: 40 kN (T), FC: 45.83 kN (T), HC: 4.167 kN (T)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δmax = 29.63 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 = 4 N/mm, L = 8000 mm
EI = 7.2 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·4·8000⁴/(384·7.2 × 10¹²) = 29.63 mm

Answer

δmax = 29.63 mm
8.Beam deflection

Answer: δmax = 4.941 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 = 30000 N, w = 15 N/mm, L = 1500 mm, EI = 6.27 × 10¹¹ N·mm²

Each load on its own

δP = PL³/(48EI) = 30000·1500³/(48·6.27 × 10¹¹) = 3.364 mm
δw = 5wL⁴/(384EI) = 5·15·1500⁴/(384·6.27 × 10¹¹) = 1.577 mm

Add them

δmax = 3.364 + 1.577 = 4.941 mm

Answer

δmax = 4.941 mm
9.Column buckling

Answer: Pcr = 40.04 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/1400² = 40043 N = 40.04 kN

Answer

Pcr = 40.04 kN
10.Column buckling

Answer: σcr = 44.76 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.7·3000 = 2100 mm
KL/r = 2100/10 = 210

Critical stress

σcr = π²E/(KL/r)² = π²·200000/210² = 44.76 MPa

Answer

σcr = 44.76 MPa

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