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
P = 25 kNM = 30 kN·mABRA​ = ?6 m1 m1 m

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 7 m from A and a counter-clockwise couple M at 6 m from A. Find the vertical reaction RA, taking upward as positive.

L = 8 mP = 25 kN at 7 mM = 30 kN·m (counter-clockwise) at 6 m
RA = kN
2.Beam support reactionsHard
P = 20 kN, 50°Q = 30 kNABα = ?1 m1.5 m2.5 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 1 m from A, inclined at 50° to the horizontal, and a vertical load Q acts at 2.5 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.

L = 5 mP = 20 kN at 1 m, θ = 50°Q = 30 kN at 2.5 m
α = °
3.Shear force & bending momentMedium
w = 10 kN/mABsection8 m2 m

Find the bending moment Mx at the section x = 8 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 10 kN/m from 0 to 10 mx = 8 m
Mx = kN·m
4.Shear force & bending momentHard
w = 24 kN/mAB2 m2 m1 m

The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x0 from A to the section where the bending moment is largest.

L = 5 mw = 24 kN/m from 2 to 4 m
x0 = m
5.Truss: method of jointsMedium
ABCDEFGH35 kN40 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.

D: 35 kN ↓G: 40 kN ↓H: 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 BG
  13. 13member DG
6.Truss: method of jointsHard
ABCDEFGH10 kN25 kN40 kN25 kN30 kN4 × 4 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: 10 kN ↓C: 25 kN ↓F: 40 kN ↓F: 25 kN →G: 30 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 = 22 kNABδ6 m2 m

A simply supported beam carries a point load P at a = 6 m from support A. Find the deflection δP under the load.

P = 22 kNa = 6 mL = 8 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 15 kN/mAB6 m

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction RB at the prop.

L = 6 mw = 15 kN/m
RB = kN
9.Column bucklingMedium
PL = 6 m25 × 80 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 = 6 mE = 200 GPa (steel)b × h = 25 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 145 kNL = 5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 145 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 145 kNn = 2.5L = 5 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 6.875 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.

Moments about A

Counter-clockwise positive.

ΣMA = 0: RB·8 − 25·7 + 30 = 0
RB = 145/8 = 18.13 kN

Vertical equilibrium

RA = 25 − 18.13 = 6.875 kN

Answer

RA = 6.875 kN

A negative reaction acts downward.

2.Beam support reactions

Answer: α = 64.75 °

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 = 20·sin 50° = 15.32 kN
Ph = 20·cos 50° = 12.86 kN

Reactions

ΣMA = 0: RB·5 = 15.32·1 + 30·2.5 → RB = 18.06 kN
Ay = 15.32 + 30 − 18.06 = 27.26 kN
Ax = Ph = 12.86 kN

Direction at A

α = atan(Ay/Ax) = atan(27.26/12.86) = 64.75°

Answer

α = 64.75°
3.Shear force & bending moment

Answer: Mx = 80 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 = 10·10 = 100 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 10·10·5
RB = 500/10 = 50 kN
ΣFy = 0: RA = 100 − 50 = 50 kN
w = 10 kN/mABRA​ = 50RB​ = 50section8 m2 m

Cut at x = 8 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 = 50·8 − 10·8·4 = 80 kN·m

Answer

Mx = 80 kN·m
4.Shear force & bending moment

Answer: x0 = 2.8 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 = 24·2 = 48 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 24·2·3
RB = 144/5 = 28.8 kN
ΣFy = 0: RA = 48 − 28.8 = 19.2 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 − 2) = 0
x0 = 2 + 19.2/24 = 2.8 m

Answer

x0 = 2.8 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: 23.44 kN (T), BC: 46.88 kN (T), CD: 46.88 kN (T), DE: 40.31 kN (T), FG: 23.44 kN (C), GH: 40.31 kN (C), AF: 39.06 kN (C), HE: 67.19 kN (C), BF: 31.25 kN (T), CG: 0, DH: 43.75 kN (T), BG: 39.06 kN (C), DG: 10.94 kN (C)

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: 83.75 kN (T), BC: 83.75 kN (T), CD: 46.25 kN (T), DE: 46.25 kN (T), FG: 92.5 kN (C), GH: 92.5 kN (C), AF: 83.09 kN (C), HE: 65.41 kN (C), BF: 10 kN (T), CG: 30 kN (C), DH: 0, FC: 12.37 kN (T), HC: 65.41 kN (T)

Answer

Zero-force member: DH
7.Beam deflection

Answer: δP = 77.65 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 = 22000 N, a = 6000 mm, b = 2000 mm, L = 8000 mm
EI = 1.7 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 22000·6000²·2000²/(3·1.7 × 10¹²·8000) = 77.65 mm

Answer

δP = 77.65 mm
8.Beam deflection

Answer: RB = 33.75 kN

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). For a statically indeterminate beam, remove a redundant support and require the deflection there to be zero (compatibility).

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δload = wL⁴/(8EI)

Compatibility at the prop

The prop force RB alone lifts the free end by RBL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

RB·L³/(3EI) = wL⁴/(8EI) → RB = 3wL/8 = 3·15·6/8 = 33.75 kN

Answer

RB = 33.75 kN
9.Column buckling

Answer: Pcr = 5.712 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·104167/6000² = 5712 N = 5.712 kN

Answer

Pcr = 5.712 kN
10.Column buckling

Answer: d = 98.34 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 = 2.5·145000 = 362500 N
KL = 1·5000 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·362500·5000²/(π³·200000))^(1/4) = 98.34 mm

Answer

d = 98.34 mm

Check: the slenderness KL/r is 203.4, so the column is slender and Euler applies.

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