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
w0​ = 4 kN/mP = 15 kNABRB​ = ?2.5 m2.5 m

The simply supported beam (pin at A, roller at B) carries a linearly varying load that rises from zero at A to 4 kN/m at B. A point load P also acts at 2.5 m from A. Find the vertical support reaction RB.

L = 5 mw0 = 4 kN/mP = 15 kN at 2.5 m
RB = kN
2.Beam support reactionsHard
w = 3 kN/mP = 20 kNABCRA​ = ?4 m3 m1 m2 m

The beam has a pin at A and a roller at B, 8 m from A, and overhangs B to the free end C. It carries a point load P at 7 m from A and a uniformly distributed load w from 4 m to the free end at 10 m. Find the vertical reaction RA, taking upward as positive.

AB = 8 mAC = 10 mP = 20 kN at 7 mw = 3 kN/m from 4 to 10 m
RA = kN
3.Shear force & bending momentMedium
w = 4 kN/mP = 20 kNABsection1 m1 m

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2 m from A. Find the bending moment Mx at the section x = 1 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 2 mP = 20 kN at 2 mw = 4 kN/m from 0 to 2 mx = 1 m
Mx = kN·m
4.Shear force & bending momentHard
w = 12 kN/mAB1 m2 m1 m

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.

L = 4 mw = 12 kN/m from 1 to 3 m
Mmax = kN·m
5.Truss: method of jointsMedium
ABBCACACB20 kN5 m2 m2 m

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

span = 5 mheight = 2 mB is 2 m from Avertical load = 20 kN
FAC = kN
6.Truss: method of jointsHard
ABBCACACB10 kN25 kN6 m1 m3 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 = 3 mB is 1 m from Avertical load = 10 kNhorizontal load = 25 kN ←
FBC = kN
7.Beam deflectionMedium
P = 8 kNABδ7 m1 m

A simply supported beam carries a point load P at a = 7 m from support A. Find the deflection δP under the load.

P = 8 kNa = 7 mL = 8 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
P = 26 kNP = 26 kNABδ1.333 m1.334 m1.333 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 = 26 kNL = 4 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 3 m30 × 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 = 3 mE = 70 GPa (aluminium)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 25 mmcross-section

A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 25 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RB = 14.17 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 triangular load

W = ½·4·5 = 10 kN
x̄ = 2L/3 = 3.333 m

Equilibrium

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

ΣMA = 0: RB·5 = 10·3.333 + 15·2.5
RB = 14.17 kN
RA = 25 − 14.17 = 10.83 kN

Answer

RB = 14.17 kN
2.Beam support reactions

Answer: RA = 4.75 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 = 3·6 = 18 kN, x̄ = (4 + 10)/2 = 7 m

Equilibrium

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

ΣMA = 0: RB·8 = 20·7 + 18·7
RB = 266/8 = 33.25 kN
RA = 38 − 33.25 = 4.75 kN

Answer

RA = 4.75 kN
3.Shear force & bending moment

Answer: Mx = −22 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.

Keep the free-end part

P: −20·1 = −20 kN·m
w: −4·1²/2 = −2 kN·m
Mx = −20 − 2 = −22 kN·m

Answer

Mx = −22 kN·m

Negative means hogging: tension on the top face.

4.Shear force & bending moment

Answer: Mmax = 18 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·2 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 12·2·2
RB = 48/4 = 12 kN
ΣFy = 0: RA = 24 − 12 = 12 kN

Zero shear

The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.

RA − w·(x0 − 1) = 0
x0 = 1 + 12/12 = 2 m

Maximum moment

Mmax = RA·x0 − w·(x0 − 1)²/2 = 12·2 − 12·1²/2 = 18 kN·m

Answer

Mmax = 18 kN·m
5.Truss: method of joints

Answer: FAC = 12 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.

ABBCACACB20 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·5 = 20·2 → Cy = 8 kN
ΣFy = 0: Ay = 20 − 8 = 12 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
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−14.42)·0.8321 = 12 kN

Answer

FAC = 12 kN (tension)
6.Truss: method of joints

Answer: FBC = 21.06 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 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·6 = 10·1 − 25·3 → Cy = −10.83 kN
ΣFy = 0: Ay = 10 − −10.83 = 20.83 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 = 3/5.831 = 0.5145, cos γ = 5/5.831 = 0.8575
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −−10.83/0.5145 = 21.06 kN

Answer

FBC = 21.06 kN (tension)
7.Beam deflection

Answer: δP = 10.14 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 = 8000 N, a = 7000 mm, b = 1000 mm, L = 8000 mm
EI = 1.61 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 8000·7000²·1000²/(3·1.61 × 10¹²·8000) = 10.14 mm

Answer

δP = 10.14 mm
8.Beam deflection

Answer: δmax = 63.92 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 = 26000 N, L = 4000 mm
EI = 9.24 × 10¹¹ N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 1333 mm
δ₁ = 31.96 mm

Add both loads

δmax = 2·δ₁ = 23PL³/(648EI) = 63.92 mm

Answer

δmax = 63.92 mm
9.Column buckling

Answer: Pcr = 3.454 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·180000/6000² = 3454 N = 3.454 kN

Answer

Pcr = 3.454 kN
10.Column buckling

Answer: σcr = 9.835 MPa

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·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.7·4000 = 2800 mm
KL/r = 2800/6.25 = 448

Critical stress

σcr = π²E/(KL/r)² = π²·200000/448² = 9.835 MPa

Answer

σcr = 9.835 MPa

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