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 = 6 kN/mP = 30 kNABMA​ = ?3 m1 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 3 m and a point load P at 4 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 4 mP = 30 kN at 4 mw = 6 kN/m from 0 to 3 m
MA = kN·m
2.Beam support reactionsHard
w = 10 kN/mP = 20 kNABRA​ = ?RB​1.5 m0.5 m2 m

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

L = 4 mP = 20 kN at 1.5 mw = 10 kN/m from 2 to 4 m
RA = kN
3.Shear force & bending momentMedium
M0​ = 35 kN·mAB1 m9 m

The simply supported beam carries a clockwise couple M0 = 35 kN·m at x = 1 m from A, and no other load. Find the bending moment Mx just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM0 = 35 kN·m clockwise, at 1 m
Mx = kN·m
4.Shear force & bending momentHard
50 kN20 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.

50 kN @ 0 m20 kN @ 12 m15 kN @ 6 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
ABBCACACB35 kN6 m5 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 5 m from Avertical load = 35 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB45 kN25 kN5 m2 m2 m

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

span = 5 mheight = 2 mB is 2 m from Avertical load = 45 kNhorizontal load = 25 kN ←
FBC = kN
7.Beam deflectionMedium
w = 10 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 = 10 kN/mL = 3 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 6 kN/mδL = 3 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 = 6 kN/mL = 3 mE = 200 GPa (steel)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 4 md = 80 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 = 4 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 120 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 120 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 120 kNn = 2L = 4 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: MA = 147 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 = 6·3 = 18 kN, x̄ = 1.5 m

Moment about the wall

MA = P·a + W·x̄ = 30·4 + 18·1.5 = 147 kN·m

Answer

MA = 147 kN·m
2.Beam support reactions

Answer: RA = 17.5 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.

w = 10 kN/mP = 20 kNABRA​RB​1.5 m0.5 m2 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 = 10·2 = 20 kN, acting at x̄ = 3 m
ΣMA = 0: RB·4 = 20·1.5 + 10·2·3
RB = 90/4 = 22.5 kN
ΣFy = 0: RA = 40 − 22.5 = 17.5 kN

Answer

RA = 17.5 kN
3.Shear force & bending moment

Answer: Mx = −3.5 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 = 35/10 = 3.5 kN (upward)
RA = −RB = −3.5 kN (downward)

Cut just left of the couple

The left part contains only RA.

Mx = RA·a = −3.5·1 = −3.5 kN·m

Answer

Mx = −3.5 kN·m
4.Shear force & bending moment

Answer: B (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 = 65 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 = −100 kN·m, MC = −70, MD = −40, ME = −40, MF = −40, MG = 0

Answer

Mark B: MB = −100 kN·m
5.Truss: method of joints

Answer: FAB = −15.71 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 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 35·5 → Cy = 29.17 kN
ΣFy = 0: Ay = 35 − 29.17 = 5.833 kN

Joint A

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

sin α = 2/5.385 = 0.3714
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −5.833/0.3714 = −15.71 kN

Answer

FAB = −15.71 kN (compression)
6.Truss: method of joints

Answer: FBC = −14.42 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 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·5 = 45·2 − 25·2 → Cy = 8 kN
ΣFy = 0: Ay = 45 − 8 = 37 kN
ΣFx = 0: Ax = 25 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 = 2/3.606 = 0.5547, cos γ = 3/3.606 = 0.8321
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −8/0.5547 = −14.42 kN

Answer

FBC = −14.42 kN (compression)
7.Beam deflection

Answer: δ = 3.164 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 = 10 N/mm, L = 3000 mm
EI = 3.2 × 10¹³ N·mm²

Deflection

δ = wL⁴/(8EI) = 10·3000⁴/(8·3.2 × 10¹³) = 3.164 mm

Answer

δ = 3.164 mm
8.Beam deflection

Answer: Ireq = 3.797 × 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 = 3000/360 = 8.333 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·6·3000⁴/(384·200000·8.333) = 3.797 × 10⁶ mm⁴
Ireq = 3.797 × 10⁶ mm⁴

Answer

Ireq = 3.797 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 21.7 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/8000² = 21704 N = 21.7 kN

Answer

Pcr = 21.7 kN
10.Column buckling

Answer: d = 86.31 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 = 2·120000 = 240000 N
KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·240000·2800²/(π³·70000))^(1/4) = 86.31 mm

Answer

d = 86.31 mm

Check: the slenderness KL/r is 129.8, so the column is slender and Euler applies.

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