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 = 35 kNM = 50 kN·mABRA​ = ?2 m6 m2 m

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 8 m from A and a clockwise couple M at 2 m from A. Find the vertical reaction RA, taking upward as positive.

L = 10 mP = 35 kN at 8 mM = 50 kN·m (clockwise) at 2 m
RA = kN
2.Beam support reactionsHard
w = 2 kN/mP = 40 kNABRA​ = ?RB​2.5 m0.5 m1 m1 m

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

L = 5 mP = 40 kN at 2.5 mw = 2 kN/m from 3 to 4 m
RA = kN
3.Shear force & bending momentMedium
10 kN55 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.

10 kN @ 2 m55 kN @ 4 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
12 kN/m40 kN45 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.

40 kN @ 0 m45 kN @ 12 m12 kN/m, 2–8 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)
5.Truss: method of jointsMedium
ABBCACACB55 kN8 m6 m3 m

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

span = 8 mheight = 3 mB is 6 m from Avertical load = 55 kN
FAC = kN
6.Truss: method of jointsHard
ABBCACACB30 kN10 kN4 m1 m3 m

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

span = 4 mheight = 3 mB is 1 m from Avertical load = 30 kNhorizontal load = 10 kN →
FAC = kN
7.Beam deflectionMedium
w = 20 kN/mδL = 3 m

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 20 kN/mL = 3 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 8 kN/mδL = 4 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 = 8 kN/mL = 4 mE = 70 GPa (aluminium)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 1.5 m25 × 60 mmcross-section

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

L = 1.5 mE = 200 GPa (steel)b × h = 25 × 60 mm
Pcr = kN
10.Column bucklingHard
P = 70 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 70 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 70 kNn = 2L = 2.5 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 2 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·10 − 35·8 − 50 = 0
RB = 330/10 = 33 kN

Vertical equilibrium

RA = 35 − 33 = 2 kN

Answer

RA = 2 kN

A negative reaction acts downward.

2.Beam support reactions

Answer: RA = 20.6 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 = 2 kN/mP = 40 kNABRA​RB​2.5 m0.5 m1 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 = 2·1 = 2 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·5 = 40·2.5 + 2·1·3.5
RB = 107/5 = 21.4 kN
ΣFy = 0: RA = 42 − 21.4 = 20.6 kN

Answer

RA = 20.6 kN
3.Shear force & bending moment

Answer: C (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 kN, R2 = 20 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 = 90, MC = 160 kN·m, MD = 120, ME = 80, MF = 40, MG = 0

Answer

Mark C: MC = 160 kN·m
4.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 = 59.33 kN, R2 = 97.67 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 = −80, MC = −65.33, MD = −98.67, ME = −180 kN·m, MF = −90, MG = 0

Answer

Mark E: ME = −180 kN·m
5.Truss: method of joints

Answer: FAC = 27.5 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.

ABBCACACB55 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·8 = 55·6 → Cy = 41.25 kN
ΣFy = 0: Ay = 55 − 41.25 = 13.75 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 = 3/3.606 = 0.8321, cos γ = 2/3.606 = 0.5547
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −41.25/0.8321 = −49.58 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−49.58)·0.5547 = 27.5 kN

Answer

FAC = 27.5 kN (tension)
6.Truss: method of joints

Answer: FAC = 15 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 kN10 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 30·1 + 10·3 → Cy = 15 kN
ΣFy = 0: Ay = 30 − 15 = 15 kN
ΣFx = 0: Ax = −10 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 = 3/4.243 = 0.7071, cos γ = 3/4.243 = 0.7071
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −15/0.7071 = −21.21 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−21.21)·0.7071 = 15 kN

Answer

FAC = 15 kN (tension)
7.Beam deflection

Answer: δ = 44.02 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 = 3000 mm
EI = 4.6 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 20·3000⁴/(8·4.6 × 10¹²) = 44.02 mm

Answer

δ = 44.02 mm
8.Beam deflection

Answer: Ireq = 34.29 × 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 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·8·4000⁴/(384·70000·11.11) = 3.429 × 10⁷ mm⁴
Ireq = 34.29 × 10⁶ mm⁴

Answer

Ireq = 34.29 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 274.2 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 0.5·1500 = 750 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·78125/750² = 274156 N = 274.2 kN

Answer

Pcr = 274.2 kN
10.Column buckling

Answer: d = 71.27 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 = 2·70000 = 140000 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·140000·2500²/(π³·70000))^(1/4) = 71.27 mm

Answer

d = 71.27 mm

Check: the slenderness KL/r is 140.3, so the column is slender and Euler applies.

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