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 = 5 kN/mP = 35 kNABRA​ = ?4.5 m0.5 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 4.5 m from the wall. Find the vertical reaction RA at the wall.

L = 5 mP = 35 kN at 4.5 mw = 5 kN/m from 0 to 5 m
RA = kN
2.Beam support reactionsHard
w = 10 kN/mP = 30 kNABCRA​ = ?1.5 m1.5 m1 m

The beam has a pin at A and a roller at B, 3 m from A, and overhangs B to the free end C. It carries a point load P at 1.5 m from A and a uniformly distributed load w from 3 m to the free end at 4 m. Find the vertical reaction RA, taking upward as positive.

AB = 3 mAC = 4 mP = 30 kN at 1.5 mw = 10 kN/m from 3 to 4 m
RA = kN
3.Shear force & bending momentMedium
w = 4 kN/mABsection2 m6 m

Find the shear force Vx at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 4 kN/m from 0 to 8 mx = 2 m
Vx = kN
4.Shear force & bending momentHard
w = 8 kN/mP = 60 kNABsection5 m1 m

Find the bending moment Mx at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 60 kN at 5 mw = 8 kN/m from 0 to 6 mx = 5 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB40 kN4 m3 m2 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 = 2 mB is 3 m from Avertical load = 40 kN
FAC = kN
6.Truss: method of jointsHard
ABBCACACB35 kN30 kN6 m5 m2 m

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

span = 6 mheight = 2 mB is 5 m from Avertical load = 35 kNhorizontal load = 30 kN ←
FAC = kN
7.Beam deflectionMedium
w = 8 kN/mP = 9 kNABδ2.5 m

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 9 kNw = 8 kN/mL = 2.5 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
P = 14 kNP = 14 kNABδ2.667 m2.666 m2.667 m

A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δmax.

P = 14 kNL = 8 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 5 m50 × 100 mmcross-section

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

L = 5 mE = 200 GPa (steel)b × h = 50 × 100 mm
Pcr = kN
10.Column bucklingHard
P = 100 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 100 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 100 kNn = 3L = 2.5 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 60 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 = 5·5 = 25 kN, x̄ = 2.5 m

Vertical equilibrium

RA = P + W = 35 + 25 = 60 kN

Answer

RA = 60 kN
2.Beam support reactions

Answer: RA = 13.33 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.

Resultant of the distributed load

W = 10·1 = 10 kN, x̄ = (3 + 4)/2 = 3.5 m

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·3 = 30·1.5 + 10·3.5
RB = 80/3 = 26.67 kN
RA = 40 − 26.67 = 13.33 kN

Answer

RA = 13.33 kN
3.Shear force & bending moment

Answer: Vx = 8 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 = 4·8 = 32 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 4·8·4
RB = 128/8 = 16 kN
ΣFy = 0: RA = 32 − 16 = 16 kN
w = 4 kN/mABRA​ = 16RB​ = 16section2 m6 m

Cut at x = 2 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 = 16 − 4·2 = 8 kN

Answer

Vx = 8 kN
4.Shear force & bending moment

Answer: Mx = 70 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·6 = 48 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 60·5 + 8·6·3
RB = 444/6 = 74 kN
ΣFy = 0: RA = 108 − 74 = 34 kN
w = 8 kN/mP = 60 kNABRA​ = 34RB​ = 74section5 m1 m

Cut at x = 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.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 34·5 − 8·5·2.5 = 70 kN·m

Answer

Mx = 70 kN·m
5.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.

ABBCACACB40 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·4 = 40·3 → Cy = 30 kN
ΣFy = 0: Ay = 40 − 30 = 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −30/0.8944 = −33.54 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−33.54)·0.4472 = 15 kN

Answer

FAC = 15 kN (tension)
6.Truss: method of joints

Answer: FAC = 9.583 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 − 30·2 → Cy = 19.17 kN
ΣFy = 0: Ay = 35 − 19.17 = 15.83 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −19.17/0.8944 = −21.43 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−21.43)·0.4472 = 9.583 kN

Answer

FAC = 9.583 kN (tension)
7.Beam deflection

Answer: δ = 7.538 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 = 9000 N, w = 8 N/mm, L = 2500 mm
EI = 1.14 × 10¹³ N·mm²

Each load on its own

δP = PL³/(3EI) = 9000·2500³/(3·1.14 × 10¹³) = 4.112 mm
δw = wL⁴/(8EI) = 8·2500⁴/(8·1.14 × 10¹³) = 3.427 mm

Add them

δ = 4.112 + 3.427 = 7.538 mm

Answer

δ = 7.538 mm
8.Beam deflection

Answer: δmax = 22.72 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 = 14000 N, L = 8000 mm
EI = 1.12 × 10¹³ N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 2667 mm
δ₁ = 11.36 mm

Add both loads

δmax = 2·δ₁ = 23PL³/(648EI) = 22.72 mm

Answer

δmax = 22.72 mm
9.Column buckling

Answer: Pcr = 20.56 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 = 100·50³/12 = 1.042 × 10⁶ mm⁴

Effective length

KL = 2·5000 = 10000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·1.042 × 10⁶/10000² = 20562 N = 20.56 kN

Answer

Pcr = 20.56 kN
10.Column buckling

Answer: d = 93.8 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 = 3·100000 = 300000 N
KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·5000²/(π³·200000))^(1/4) = 93.8 mm

Answer

d = 93.8 mm

Check: the slenderness KL/r is 213.2, so the column is slender and Euler applies.

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