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 = 8 kN/mP = 10 kNABRA​ = ?2.5 m2.5 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 2.5 m from the wall. Find the vertical reaction RA at the wall.

L = 5 mP = 10 kN at 2.5 mw = 8 kN/m from 0 to 5 m
RA = kN
2.Beam support reactionsHard
P = 50 kNQ = 75 kNADCRD​ = ?2 m2 m1 m3 m

A compound beam is fixed to the wall at A, has an internal hinge at C (4 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 5 m from A and a point load Q at 2 m from A. Find the reaction RD at the roller.

AC = 4 mAD = 8 mP = 50 kN at 5 mQ = 75 kN at 2 m
RD = kN
3.Shear force & bending momentMedium
25 kN15 kNABCDEFG2 m2 m2 m2 m2 m2 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.

25 kN @ 8 m15 kN @ 2 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
4.Shear force & bending momentHard
15 kN/m20 kN50 kN45 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.

20 kN @ 0 m50 kN @ 6 m45 kN @ 2 m15 kN/m, 1–4 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
ABBCACACB30 kN8 m6 m4 m

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

span = 8 mheight = 4 mB is 6 m from Avertical load = 30 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB35 kN30 kN6 m5 m4 m

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

span = 6 mheight = 4 mB is 5 m from Avertical load = 35 kNhorizontal load = 30 kN ←
FBC = kN
7.Beam deflectionMedium
w = 5 kN/mABθ6 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.

w = 5 kN/mL = 6 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴
θA = rad
8.Beam deflectionHard
P = 5 kNABδ3 m1 m

A simply supported beam carries a point load P at b = 1 m from support B. Find the maximum deflection δmax of the beam.

P = 5 kNb = 1 mL = 4 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 1.5 md = 60 mmcross-section

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 100 kNL = 3 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 100 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 100 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 50 kN

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 = 8·5 = 40 kN, x̄ = 2.5 m

Vertical equilibrium

RA = P + W = 10 + 40 = 50 kN

Answer

RA = 50 kN
2.Beam support reactions

Answer: RD = 12.5 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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

ΣMC = 0: RD·4 = 50·1
RD = 12.5 kN
VC = 50 − 12.5 = 37.5 kN

Answer

RD = 12.5 kN
3.Shear force & bending moment

Answer: E (x = 8 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 = 20.83 kN, R2 = 19.17 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 = 41.67, MC = 53.33, MD = 65, ME = 76.67 kN·m, MF = 38.33, MG = 0

Answer

Mark E: ME = 76.67 kN·m
4.Shear force & bending moment

Answer: E (x = 4 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 = 45.83 kN, R2 = 114.2 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 = −20, MC = −1.667, MD = −43.33, ME = −100 kN·m, MF = −50, MG = 0

Answer

Mark E: ME = −100 kN·m
5.Truss: method of joints

Answer: FAB = −13.52 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.

ABBCACACB30 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·8 = 30·6 → Cy = 22.5 kN
ΣFy = 0: Ay = 30 − 22.5 = 7.5 kN

Joint A

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

sin α = 4/7.211 = 0.5547
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −7.5/0.5547 = −13.52 kN

Answer

FAB = −13.52 kN (compression)
6.Truss: method of joints

Answer: FBC = −9.449 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.

ABBCACACB35 kN30 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·6 = 35·5 − 30·4 → Cy = 9.167 kN
ΣFy = 0: Ay = 35 − 9.167 = 25.83 kN
ΣFx = 0: Ax = 30 kN

Joint C

The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.

sin γ = 4/4.123 = 0.9701, cos γ = 1/4.123 = 0.2425
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −9.167/0.9701 = −9.449 kN

Answer

FBC = −9.449 kN (compression)
7.Beam deflection

Answer: θA = 0.003947 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 5 N/mm, L = 6000 mm
EI = 1.14 × 10¹³ N·mm²

End slope

θA = wL³/(24EI) = 5·6000³/(24·1.14 × 10¹³) = 0.003947 rad

Answer

θA = 0.003947 rad

That is 0.2262°.

8.Beam deflection

Answer: δmax = 2.893 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 = 5000 N, b = 1000 mm, L = 4000 mm
EI = 1.61 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((4000² − 1000²)/3) = 2236 mm

Maximum deflection

δmax = Pb(L² − b²)3/2/(9√3·EIL) = 2.893 mm

Answer

δmax = 2.893 mm
9.Column buckling

Answer: Pcr = 48.83 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 = 2·1500 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/3000² = 48835 N = 48.83 kN

Answer

Pcr = 48.83 kN
10.Column buckling

Answer: d = 85.35 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 = 2·100000 = 200000 N
KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·3000²/(π³·70000))^(1/4) = 85.35 mm

Answer

d = 85.35 mm

Check: the slenderness KL/r is 140.6, so the column is slender and Euler applies.

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