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
P1​ = 20 kNP2​ = 15 kNABRA​RB​ = ?1 m5 m2 m

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

L = 8 mP1 = 20 kN at 1 mP2 = 15 kN at 6 m
RB = kN
2.Beam support reactionsHard
w = 12 kN/mP = 55 kNABRA​RB​ = ?3 m2 m1 m

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

L = 6 mP = 55 kN at 5 mw = 12 kN/m from 3 to 5 m
RB = kN
3.Shear force & bending momentMedium
w = 6 kN/mABsection1.5 m2.5 m

Find the shear force Vx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 6 kN/m from 0 to 4 mx = 1.5 m
Vx = kN
4.Shear force & bending momentHard
w = 6 kN/mP = 55 kNABsection2.5 m1.5 m

Find the bending moment Mx at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 55 kN at 2.5 mw = 6 kN/m from 0 to 4 mx = 2.5 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB40 kN8 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 = 8 mheight = 2 mB is 2 m from Avertical load = 40 kN
FBC = kN
6.Truss: method of jointsHard
ABBCACACB40 kN5 kN5 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 = 5 mheight = 3 mB is 1 m from Avertical load = 40 kNhorizontal load = 5 kN →
FBC = kN
7.Beam deflectionMedium
P = 24 kNABδ4.5 m1.5 m

A simply supported beam carries a point load P at a = 4.5 m from support A. Find the deflection δP under the load.

P = 24 kNa = 4.5 mL = 6 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 6 kN/mδL = 1.5 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 = 1.5 mE = 70 GPa (aluminium)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 1.5 m20 × 100 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)b × h = 20 × 100 mm
Pcr = kN
10.Column bucklingHard
P = 135 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 135 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 135 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 13.75 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.

P1​ = 20 kNP2​ = 15 kNABRA​RB​1 m5 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.

ΣMA = 0: RB·8 = 20·1 + 15·6
RB = 110/8 = 13.75 kN
ΣFy = 0: RA = 35 − 13.75 = 21.25 kN

Answer

RB = 13.75 kN
2.Beam support reactions

Answer: RB = 61.83 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 = 12 kN/mP = 55 kNABRA​RB​3 m2 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 = 12·2 = 24 kN, acting at x̄ = 4 m
ΣMA = 0: RB·6 = 55·5 + 12·2·4
RB = 371/6 = 61.83 kN
ΣFy = 0: RA = 79 − 61.83 = 17.17 kN

Answer

RB = 61.83 kN
3.Shear force & bending moment

Answer: Vx = 3 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 = 6·4 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 6·4·2
RB = 48/4 = 12 kN
ΣFy = 0: RA = 24 − 12 = 12 kN
w = 6 kN/mABRA​ = 12RB​ = 12section1.5 m2.5 m

Cut at x = 1.5 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.

V = RA − (all downward loads left of the cut)
Vx = 12 − 6·1.5 = 3 kN

Answer

Vx = 3 kN
4.Shear force & bending moment

Answer: Mx = 62.81 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 = 6·4 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 55·2.5 + 6·4·2
RB = 185.5/4 = 46.38 kN
ΣFy = 0: RA = 79 − 46.38 = 32.63 kN
w = 6 kN/mP = 55 kNABRA​ = 32.63RB​ = 46.38section2.5 m1.5 m

Cut at x = 2.5 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 = 32.63·2.5 − 6·2.5·1.25 = 62.81 kN·m

Answer

Mx = 62.81 kN·m
5.Truss: method of joints

Answer: FBC = −31.62 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·8 = 40·2 → Cy = 10 kN
ΣFy = 0: Ay = 40 − 10 = 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 = 2/6.325 = 0.3162, cos γ = 6/6.325 = 0.9487
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −10/0.3162 = −31.62 kN

Answer

FBC = −31.62 kN (compression)
6.Truss: method of joints

Answer: FBC = −18.33 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 kN5 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·5 = 40·1 + 5·3 → Cy = 11 kN
ΣFy = 0: Ay = 40 − 11 = 29 kN
ΣFx = 0: Ax = −5 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 = 0.6, cos γ = 4/5 = 0.8
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −11/0.6 = −18.33 kN

Answer

FBC = −18.33 kN (compression)
7.Beam deflection

Answer: δP = 35.74 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 = 24000 N, a = 4500 mm, b = 1500 mm, L = 6000 mm
EI = 1.7 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 24000·4500²·1500²/(3·1.7 × 10¹²·6000) = 35.74 mm

Answer

δP = 35.74 mm
8.Beam deflection

Answer: Ireq = 1.356 × 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 = 1500/360 = 4.167 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·6·1500⁴/(384·70000·4.167) = 1.356 × 10⁶ mm⁴
Ireq = 1.356 × 10⁶ mm⁴

Answer

Ireq = 1.356 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 5.118 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 = 100·20³/12 = 66667 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·66667/3000² = 5118 N = 5.118 kN

Answer

Pcr = 5.118 kN
10.Column buckling

Answer: d = 68.31 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 = 2.5·135000 = 337500 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·337500·2500²/(π³·200000))^(1/4) = 68.31 mm

Answer

d = 68.31 mm

Check: the slenderness KL/r is 146.4, so the column is slender and Euler applies.

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