Structural Analysis: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
w = 12 kN/mP = 10 kNABMA​ = ?3 m7 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 3 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 10 mP = 10 kN at 3 mw = 12 kN/m from 0 to 10 m
MA = kN·m
2.Beam support reactionsHard
w = 12 kN/mP = 50 kNABCRA​ = ?3 m0.5 m1 m1.5 m

The beam has a pin at A and a roller at B, 4.5 m from A, and overhangs B to the free end C. It carries a point load P at 3.5 m from A and a uniformly distributed load w from 3 m to the free end at 6 m. Find the vertical reaction RA, taking upward as positive.

AB = 4.5 mAC = 6 mP = 50 kN at 3.5 mw = 12 kN/m from 3 to 6 m
RA = kN
3.Shear force & bending momentMedium
w = 5 kN/mP = 10 kNAB1.5 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 1.5 m from A. Find the bending moment MA at the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 10 kN at 1.5 mw = 5 kN/m from 0 to 4 m
MA = kN·m
4.Shear force & bending momentHard
w = 2 kN/mABC5 m2.5 m

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment Mmax between the supports.

L = 7.5 mw = 2 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m
Mmax = kN·m
5.Truss: method of jointsMedium
ABCDEFGH15 kN15 kN10 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: 15 kN ↓C: 15 kN ↓D: 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 FC
  13. 13member HC
6.Truss: method of jointsHard
ABBCACACB35 kN25 kN6 m2 m3 m

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

span = 6 mheight = 3 mB is 2 m from Avertical load = 35 kNhorizontal load = 25 kN →
FBC = kN
7.Beam deflectionMedium
P = 13 kNABδ3 m5 m

A simply supported beam carries a point load P at a = 3 m from support A. Find the deflection δP under the load.

P = 13 kNa = 3 mL = 8 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 5 kN/mP = 23 kNδL = 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 = 23 kNw = 5 kN/mL = 5 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 3 m30 × 40 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 = 3 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
P = 95 kNL = 2 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 95 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 95 kNn = 2L = 2 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: MA = 630 kN·m

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.

Resultant of the distributed load

W = 12·10 = 120 kN, x̄ = 5 m

Moment about the wall

MA = P·a + W·x̄ = 10·3 + 120·5 = 630 kN·m

Answer

MA = 630 kN·m
2.Beam support reactions

Answer: RA = 11.11 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.

Resultant of the distributed load

W = 12·3 = 36 kN, x̄ = (3 + 6)/2 = 4.5 m

Equilibrium

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

ΣMA = 0: RB·4.5 = 50·3.5 + 36·4.5
RB = 337/4.5 = 74.89 kN
RA = 86 − 74.89 = 11.11 kN

Answer

RA = 11.11 kN
3.Shear force & bending moment

Answer: MA = −55 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: −10·1.5 = −15 kN·m
w: −5·4²/2 = −40 kN·m
MA = −15 − 40 = −55 kN·m

Answer

MA = −55 kN·m

Negative means hogging: tension on the top face.

4.Shear force & bending moment

Answer: Mmax = 3.516 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 2·7.5 = 15 kN, acting at x̄ = 3.75 m
ΣMA = 0: RB·5 = 2·7.5·3.75
RB = 56.25/5 = 11.25 kN
ΣFy = 0: RA = 15 − 11.25 = 3.75 kN

Zero shear

RA − w·x0 = 0 → x0 = 3.75/2 = 1.875 m

Maximum sagging moment

Mmax = RA·x0 − w·x0²/2 = RA²/(2w) = 3.75²/(2·2) = 3.516 kN·m

Answer

Mmax = 3.516 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 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: 15.94 kN (T), BC: 15.94 kN (T), CD: 14.06 kN (T), DE: 14.06 kN (T), FG: 20.62 kN (C), GH: 20.62 kN (C), AF: 26.56 kN (C), HE: 23.44 kN (C), BF: 15 kN (T), CG: 0, DH: 10 kN (T), FC: 7.813 kN (T), HC: 10.94 kN (T)

Answer

Zero-force member: CG
6.Truss: method of joints

Answer: FBC = −40.28 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.

ABBCACACB35 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·6 = 35·2 + 25·3 → Cy = 24.17 kN
ΣFy = 0: Ay = 35 − 24.17 = 10.83 kN
ΣFx = 0: Ax = −25 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 = 3/5 = 0.6, cos γ = 4/5 = 0.8
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −24.17/0.6 = −40.28 kN

Answer

FBC = −40.28 kN (compression)
7.Beam deflection

Answer: δP = 69.25 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 = 13000 N, a = 3000 mm, b = 5000 mm, L = 8000 mm
EI = 1.76 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 13000·3000²·5000²/(3·1.76 × 10¹²·8000) = 69.25 mm

Answer

δP = 69.25 mm
8.Beam deflection

Answer: δmax = 8.981 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 = 23000 N, w = 5 N/mm, L = 5000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δP = PL³/(48EI) = 23000·5000³/(48·1.12 × 10¹³) = 5.348 mm
δw = 5wL⁴/(384EI) = 5·5·5000⁴/(384·1.12 × 10¹³) = 3.633 mm

Add them

δmax = 5.348 + 3.633 = 8.981 mm

Answer

δmax = 8.981 mm
9.Column buckling

Answer: Pcr = 40.28 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 = 0.7·3000 = 2100 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/2100² = 40284 N = 40.28 kN

Answer

Pcr = 40.28 kN
10.Column buckling

Answer: d = 68.8 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 = 2·95000 = 190000 N
KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·190000·2000²/(π³·70000))^(1/4) = 68.8 mm

Answer

d = 68.8 mm

Check: the slenderness KL/r is 116.3, so the column is slender and Euler applies.

ExamProblems · examproblems.com/courses/structural-analysis