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​ = 25 kNP2​ = 50 kNABRA​RB​ = ?1 m6 m3 m

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

L = 10 mP1 = 25 kN at 1 mP2 = 50 kN at 7 m
RB = kN
2.Beam support reactionsHard
w = 6 kN/mP = 15 kNABRA​RB​ = ?1 m0.5 m3.5 m1 m

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

L = 6 mP = 15 kN at 1.5 mw = 6 kN/m from 1 to 5 m
RB = kN
3.Shear force & bending momentMedium
w = 12 kN/mABsection2.5 m7.5 m

Find the shear force Vx at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 12 kN/m from 0 to 10 mx = 2.5 m
Vx = kN
4.Shear force & bending momentHard
5 kN/m10 kN30 kNABCDEFG1 m1 m1 m1 m1 m1 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

10 kN @ 6 m30 kN @ 2 m5 kN/m, 0–5 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
5.Truss: method of jointsMedium
ABBCACACB10 kN6 m2 m2 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 = 2 mB is 2 m from Avertical load = 10 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB40 kN20 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 = 40 kNhorizontal load = 20 kN →
FAB = kN
7.Beam deflectionMedium
w = 6 kN/mδL = 2.5 m

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

w = 6 kN/mL = 2.5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 6 kN/mP = 12 kNδL = 1.5 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 = 12 kNw = 6 kN/mL = 1.5 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 3 md = 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 = 3 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 85 kNL = 4 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 85 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 37.5 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​ = 25 kNP2​ = 50 kNABRA​RB​1 m6 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·10 = 25·1 + 50·7
RB = 375/10 = 37.5 kN
ΣFy = 0: RA = 75 − 37.5 = 37.5 kN

Answer

RB = 37.5 kN
2.Beam support reactions

Answer: RB = 15.75 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.

w = 6 kN/mP = 15 kNABRA​RB​1 m0.5 m3.5 m1 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.

UDL resultant: W = w·ℓ = 6·4 = 24 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 15·1.5 + 6·4·3
RB = 94.5/6 = 15.75 kN
ΣFy = 0: RA = 39 − 15.75 = 23.25 kN

Answer

RB = 15.75 kN
3.Shear force & bending moment

Answer: Vx = 30 kN

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 = 12·10 = 120 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 12·10·5
RB = 600/10 = 60 kN
ΣFy = 0: RA = 120 − 60 = 60 kN
w = 12 kN/mABRA​ = 60RB​ = 60section2.5 m7.5 m

Cut at x = 2.5 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.

V = RA − (all downward loads left of the cut)
Vx = 60 − 12·2.5 = 30 kN

Answer

Vx = 30 kN
4.Shear force & bending moment

Answer: C (x = 2 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 28.5 kN, R2 = 36.5 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 26, MC = 47 kN·m, MD = 33, ME = 14, MF = −10, MG = 0

Answer

Mark C: MC = 47 kN·m
5.Truss: method of joints

Answer: FAB = −9.428 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.

ABBCACACB10 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 10·2 → Cy = 3.333 kN
ΣFy = 0: Ay = 10 − 3.333 = 6.667 kN

Joint A

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

sin α = 2/2.828 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −6.667/0.7071 = −9.428 kN

Answer

FAB = −9.428 kN (compression)
6.Truss: method of joints

Answer: FAB = −22.36 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.

ABBCACACB40 kN20 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 40·1 + 20·2 → Cy = 20 kN
ΣFy = 0: Ay = 40 − 20 = 20 kN
ΣFx = 0: Ax = −20 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 = −20/0.8944 = −22.36 kN

Answer

FAB = −22.36 kN (compression)
7.Beam deflection

Answer: δmax = 1.795 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 = 6 N/mm, L = 2500 mm
EI = 1.7 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·6·2500⁴/(384·1.7 × 10¹²) = 1.795 mm

Answer

δmax = 1.795 mm
8.Beam deflection

Answer: δmax = 3.129 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 = 12000 N, w = 6 N/mm, L = 1500 mm, EI = 3.96 × 10¹¹ N·mm²

Each load on its own

δP = PL³/(48EI) = 12000·1500³/(48·3.96 × 10¹¹) = 2.131 mm
δw = 5wL⁴/(384EI) = 5·6·1500⁴/(384·3.96 × 10¹¹) = 0.9988 mm

Add them

δmax = 2.131 + 0.9988 = 3.129 mm

Answer

δmax = 3.129 mm
9.Column buckling

Answer: Pcr = 139.5 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·636173/3000² = 139528 N = 139.5 kN

Answer

Pcr = 139.5 kN
10.Column buckling

Answer: d = 83.72 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 = 2.5·85000 = 212500 N
KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·2800²/(π³·70000))^(1/4) = 83.72 mm

Answer

d = 83.72 mm

Check: the slenderness KL/r is 133.8, so the column is slender and Euler applies.

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