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
P1​ = 10 kNP2​ = 40 kNABRA​ = ?RB​2 m0.5 m1.5 m

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

L = 4 mP1 = 10 kN at 2 mP2 = 40 kN at 2.5 m
RA = kN
2.Beam support reactionsHard
P = 15 kN, 50°Q = 35 kNABRA​ = ?2.5 m3 m0.5 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 50° to the horizontal, and a vertical load Q acts at 5.5 m from A. Find the magnitude of the total reaction force RA at the pin A.

L = 6 mP = 15 kN at 2.5 m, θ = 50°Q = 35 kN at 5.5 m
RA = kN
3.Shear force & bending momentMedium
M0​ = 60 kN·mAB2 m3 m

The simply supported beam carries a clockwise couple M0 = 60 kN·m at x = 2 m from A, and no other load. Find the bending moment Mx just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 5 mM0 = 60 kN·m clockwise, at 2 m
Mx = kN·m
4.Shear force & bending momentHard
w = 6 kN/mAB4 m6 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 = 10 mw = 6 kN/m from 4 to 10 m
x0 = m
5.Truss: method of jointsMedium
ABBCACACB60 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 = 60 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB45 kN5 kN4 m1 m2 m

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

span = 4 mheight = 2 mB is 1 m from Avertical load = 45 kNhorizontal load = 5 kN →
FAB = kN
7.Beam deflectionMedium
w = 2 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 = 2 kN/mL = 5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 5 kN/mδL = 6 m

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.

w = 5 kN/mL = 6 mE = 70 GPa (aluminium)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 1.5 m30 × 50 mmcross-section

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

L = 1.5 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 30 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 = 30 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RA = 20 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​ = 40 kNABRA​RB​2 m0.5 m1.5 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·4 = 10·2 + 40·2.5
RB = 120/4 = 30 kN
ΣFy = 0: RA = 50 − 30 = 20 kN

Answer

RA = 20 kN
2.Beam support reactions

Answer: RA = 13.62 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 = 15·sin 50° = 11.49 kN
Ph = 15·cos 50° = 9.642 kN

Reactions

ΣMA = 0: RB·6 = 11.49·2.5 + 35·5.5 → RB = 36.87 kN
Ay = 11.49 + 35 − 36.87 = 9.62 kN
Ax = Ph = 9.642 kN

Resultant at A

RA = √(9.642² + 9.62²) = 13.62 kN

Answer

RA = 13.62 kN
3.Shear force & bending moment

Answer: Mx = 36 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.

RB = M0/L = 60/5 = 12 kN (upward)
RA = −RB = −12 kN (downward)

Cut just right of the couple

The left part now contains RA and the couple.

Mx = RA·a + M0 = −12·2 + 60 = 36 kN·m

Answer

Mx = 36 kN·m
4.Shear force & bending moment

Answer: x0 = 5.8 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 = 6·6 = 36 kN, acting at x̄ = 7 m
ΣMA = 0: RB·10 = 6·6·7
RB = 252/10 = 25.2 kN
ΣFy = 0: RA = 36 − 25.2 = 10.8 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 − 4) = 0
x0 = 4 + 10.8/6 = 5.8 m

Answer

x0 = 5.8 m
5.Truss: method of joints

Answer: FAB = −16.01 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.

ABBCACACB60 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 60·5 → Cy = 50 kN
ΣFy = 0: Ay = 60 − 50 = 10 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 = −10/0.6247 = −16.01 kN

Answer

FAB = −16.01 kN (compression)
6.Truss: method of joints

Answer: FAB = −34.94 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.

ABBCACACB45 kN5 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 45·1 + 5·2 → Cy = 13.75 kN
ΣFy = 0: Ay = 45 − 13.75 = 31.25 kN
ΣFx = 0: Ax = −5 kN

Joint A

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

sin α = 2/2.236 = 0.8944
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −31.25/0.8944 = −34.94 kN

Answer

FAB = −34.94 kN (compression)
7.Beam deflection

Answer: δmax = 25.96 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 = 2 N/mm, L = 5000 mm
EI = 6.27 × 10¹¹ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·2·5000⁴/(384·6.27 × 10¹¹) = 25.96 mm

Answer

δmax = 25.96 mm
8.Beam deflection

Answer: Ireq = 72.32 × 10⁶ mm⁴

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).

Deflection limit

δallow = L/360 = 6000/360 = 16.67 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·5·6000⁴/(384·70000·16.67) = 7.232 × 10⁷ mm⁴
Ireq = 72.32 × 10⁶ mm⁴

Answer

Ireq = 72.32 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 34.54 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·112500/1500² = 34544 N = 34.54 kN

Answer

Pcr = 34.54 kN
10.Column buckling

Answer: σcr = 12.34 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 1·3000 = 3000 mm
KL/r = 3000/7.5 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σcr = 12.34 MPa

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