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
P1​ = 10 kNP2​ = 55 kNABRA​RB​ = ?1 m1 m3 m

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

L = 5 mP1 = 10 kN at 1 mP2 = 55 kN at 2 m
RB = kN
2.Beam support reactionsHard
w0​ = 6 kN/mP = 40 kNABMA​ = ?10 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 6 kN/m at the free end B, plus a point load P at 10 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 10 mw0 = 6 kN/mP = 40 kN at 10 m
MA = kN·m
3.Shear force & bending momentMedium
P1​ = 50 kNP2​ = 30 kNABC4.5 m1.5 m2 m

The beam rests on a pin at A and a roller at B and overhangs B by 2 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 = 8 mP1 = 50 kN at 4.5 mP2 = 30 kN at 8 msupports at x = 0 and x = 6 m
Mx = kN·m
4.Shear force & bending momentHard
w = 8 kN/mP = 20 kNABsection3 m7 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 = 10 mP = 20 kN at 3 mw = 8 kN/m from 0 to 10 mx = 3 m
Mx = kN·m
5.Truss: method of jointsMedium
ABCDEFGH30 kN25 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.

D: 30 kN ↓H: 25 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 BG
  13. 13member DG
6.Truss: method of jointsHard
ABCDEFGH40 kN20 kN40 kN10 kN35 kN4 × 3 m3 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: 40 kN ↓F: 20 kN ↓F: 40 kN →G: 10 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
P = 23 kNABδ4.5 m1.5 m

A simply supported beam carries a point load P at a = 4.5 m from support A. Find the deflection δP under the load.

P = 23 kNa = 4.5 mL = 6 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 8 kN/mP = 29 kNδL = 4 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 = 29 kNw = 8 kN/mL = 4 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 2 m25 × 60 mmcross-section

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

L = 2 mE = 200 GPa (steel)b × h = 25 × 60 mm
Pcr = kN
10.Column bucklingHard
P = 105 kNL = 2 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 105 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 105 kNn = 2L = 2 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 24 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​ = 10 kNP2​ = 55 kNABRA​RB​1 m1 m3 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 = 10·1 + 55·2
RB = 120/5 = 24 kN
ΣFy = 0: RA = 65 − 24 = 41 kN

Answer

RB = 24 kN
2.Beam support reactions

Answer: MA = 600 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 triangular load

W = ½·6·10 = 30 kN
x̄ = 2L/3 = 6.667 m

Moment about the wall

MA = W·x̄ + P·a = 30·6.667 + 40·10 = 600 kN·m

Answer

MA = 600 kN·m
3.Shear force & bending moment

Answer: Mx = 11.25 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.

Support reactions

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

ΣMA = 0: RB·6 = 50·4.5 + 30·8
RB = 465/6 = 77.5 kN
ΣFy = 0: RA = 80 − 77.5 = 2.5 kN

Cut under P1 and keep the left part

Mx = RA·4.5 = 2.5·4.5 = 11.25 kN·m

Answer

Mx = 11.25 kN·m
4.Shear force & bending moment

Answer: Mx = 126 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 = 8·10 = 80 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 20·3 + 8·10·5
RB = 460/10 = 46 kN
ΣFy = 0: RA = 100 − 46 = 54 kN
w = 8 kN/mP = 20 kNABRA​ = 54RB​ = 46section3 m7 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 = 54·3 − 8·3·1.5 = 126 kN·m

Answer

Mx = 126 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 C

BC and CD 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: 10.31 kN (T), BC: 20.62 kN (T), CD: 20.62 kN (T), DE: 30.94 kN (T), FG: 10.31 kN (C), GH: 30.94 kN (C), AF: 17.19 kN (C), HE: 51.56 kN (C), BF: 13.75 kN (T), CG: 0, DH: 16.25 kN (T), BG: 17.19 kN (C), DG: 17.19 kN (T)

Answer

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

Answer: member DH

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 D

CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.

ΣF⊥ = 0 → FDH = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 88.75 kN (T), BC: 88.75 kN (T), CD: 56.25 kN (T), DE: 56.25 kN (T), FG: 77.5 kN (C), GH: 77.5 kN (C), AF: 68.94 kN (C), HE: 79.55 kN (C), BF: 40 kN (T), CG: 10 kN (C), DH: 0, FC: 15.91 kN (C), HC: 30.05 kN (T)

Answer

Zero-force member: DH
7.Beam deflection

Answer: δP = 44.11 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, a = 4500 mm, b = 1500 mm, L = 6000 mm
EI = 1.32 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 23000·4500²·1500²/(3·1.32 × 10¹²·6000) = 44.11 mm

Answer

δP = 44.11 mm
8.Beam deflection

Answer: δmax = 5.833 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 = 29000 N, w = 8 N/mm, L = 4000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δP = PL³/(48EI) = 29000·4000³/(48·1.12 × 10¹³) = 3.452 mm
δw = 5wL⁴/(384EI) = 5·8·4000⁴/(384·1.12 × 10¹³) = 2.381 mm

Add them

δmax = 3.452 + 2.381 = 5.833 mm

Answer

δmax = 5.833 mm
9.Column buckling

Answer: Pcr = 38.55 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·78125/2000² = 38553 N = 38.55 kN

Answer

Pcr = 38.55 kN
10.Column buckling

Answer: d = 70.55 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 = 2·105000 = 210000 N
KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·210000·2000²/(π³·70000))^(1/4) = 70.55 mm

Answer

d = 70.55 mm

Check: the slenderness KL/r is 113.4, so the column is slender and Euler applies.

ExamProblems · examproblems.com/courses/structural-analysis