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
w = 10 kN/mP = 35 kNABMA​ = ?2 m0.5 m2.5 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 2.5 m and a point load P at 2 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 5 mP = 35 kN at 2 mw = 10 kN/m from 0 to 2.5 m
MA = kN·m
2.Beam support reactionsHard
w0​ = 15 kN/mP = 30 kNABMA​ = ?4 m4 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 15 kN/m at the free end B, plus a point load P at 4 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 8 mw0 = 15 kN/mP = 30 kN at 4 m
MA = kN·m
3.Shear force & bending momentMedium
w = 12 kN/mP = 70 kNAB3 m2 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 MA at the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 70 kN at 3 mw = 12 kN/m from 0 to 5 m
MA = kN·m
4.Shear force & bending momentHard
w = 2 kN/mP = 30 kNABsection1 m4 m

Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 30 kN at 1 mw = 2 kN/m from 0 to 5 mx = 1 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB50 kN6 m5 m4 m

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

span = 6 mheight = 4 mB is 5 m from Avertical load = 50 kN
FAB = kN
6.Truss: method of jointsHard
ABCDEFGH15 kN35 kN30 kN4 × 4 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: 15 kN ↓F: 35 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 BG
  13. 13member DG
7.Beam deflectionMedium
w = 20 kN/mδL = 5 m

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

w = 20 kN/mL = 5 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
P = 10 kNAB3 m3 m

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the size of the bending moment MA at the fixed end.

L = 6 mP = 10 kN at 3 m
MA = kN·m
9.Column bucklingMedium
PL = 2 md = 20 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 80 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 80 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: MA = 101.3 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 = 10·2.5 = 25 kN, x̄ = 1.25 m

Moment about the wall

MA = P·a + W·x̄ = 35·2 + 25·1.25 = 101.3 kN·m

Answer

MA = 101.3 kN·m
2.Beam support reactions

Answer: MA = 440 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 = ½·15·8 = 60 kN
x̄ = 2L/3 = 5.333 m

Moment about the wall

MA = W·x̄ + P·a = 60·5.333 + 30·4 = 440 kN·m

Answer

MA = 440 kN·m
3.Shear force & bending moment

Answer: MA = −360 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: −70·3 = −210 kN·m
w: −12·5²/2 = −150 kN·m
MA = −210 − 150 = −360 kN·m

Answer

MA = −360 kN·m

Negative means hogging: tension on the top face.

4.Shear force & bending moment

Answer: Mx = 28 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 = 2·5 = 10 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 30·1 + 2·5·2.5
RB = 55/5 = 11 kN
ΣFy = 0: RA = 40 − 11 = 29 kN
w = 2 kN/mP = 30 kNABRA​ = 29RB​ = 11section1 m4 m

Cut at x = 1 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 = 29·1 − 2·1·0.5 = 28 kN·m

Answer

Mx = 28 kN·m
5.Truss: method of joints

Answer: FAB = −13.34 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.

ABBCACACB50 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 50·5 → Cy = 41.67 kN
ΣFy = 0: Ay = 50 − 41.67 = 8.333 kN

Joint A

Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.

sin α = 4/6.403 = 0.6247
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −8.333/0.6247 = −13.34 kN

Answer

FAB = −13.34 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 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: 61.25 kN (T), BC: 67.5 kN (T), CD: 67.5 kN (T), DE: 33.75 kN (T), FG: 61.25 kN (C), GH: 33.75 kN (C), AF: 32.81 kN (C), HE: 42.19 kN (C), BF: 19.69 kN (T), CG: 0, DH: 25.31 kN (T), BG: 7.813 kN (C), DG: 42.19 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δmax = 19.38 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 = 20 N/mm, L = 5000 mm
EI = 8.4 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·20·5000⁴/(384·8.4 × 10¹²) = 19.38 mm

Answer

δmax = 19.38 mm
8.Beam deflection

Answer: MA = 11.25 kN·m

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 = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → RB = 5P/16 = 5·10/16 = 3.125 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

MA = P·L/2 − RB·L = 30 − 18.75 = 11.25 kN·m

Answer

MA = 11.25 kN·m
9.Column buckling

Answer: Pcr = 5.426 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·2000 = 1000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·7854/1000² = 5426 N = 5.426 kN

Answer

Pcr = 5.426 kN
10.Column buckling

Answer: σcr = 87.73 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 1·3000 = 3000 mm
KL/r = 3000/20 = 150

Critical stress

σcr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa

Answer

σcr = 87.73 MPa

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