Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

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/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

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: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

Find the elongation δ of the brass bar under the axial load.

P = 40 kNd = 30 mmL = 1.5 mE = 100 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaP = 40 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid brass shaft 0.8 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 200 N·mL = 0.8 mG = 39 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A solid brass shaft transmits 30 kW at 1500 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 30 kWN = 1500 rpmτ_allow = 80 MPaθ_allow = 0.5°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

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 @ 5 m40 kN @ 3 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 4 mP = 35 kN at 1.5 mw = 4 kN/m from 0 to 4 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 160 mm wide with two upright legs, 80 mm high overall. Base and legs are all 14 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 160 mmh = 80 mmt = 14 mm

ȳ = mm

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 200 mm wide and 12 mm thick, an overall depth of 440 mm and a web 10 mm thick. It carries a bending moment of 40 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 200 mmH = 440 mmt_f = 12 mmt_w = 10 mmM = 40 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 20 MPa, σᵧ = 80 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 20 MPaσ_y = 80 MPaτ_xy = −60 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 750 N·m together with a torque of 500 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 40 mmM = 750 N·mT = 500 N·m

σ₁ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 0.5 m from A. Find the deflection δ_B at the free end B.

P = 17 kNa = 0.5 mL = 2.5 mE = 11 GPa (timber)I = 8.5 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 5 kN/mL = 8 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 3 mE = 70 GPa (aluminium)b × h = 40 × 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 30 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 30 kNn = 3L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.8488 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·30²/4 = 706.9 mm²

σ = P/A = 40·1000/706.9 = 56.59 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 56.59/100000 = 0.0005659 = 565.9 µε

Elongation

δ = ε·L = 0.0005659·1500 mm = 0.8488 mm

Answer

δ = 0.8488 mm

2.Axial stress & strain

Answer: σ₂ = 17.78 MPa

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Stiffnesses

k₁ = 200000·314.2/600 = 104720 N/mm

k₂ = 110000·1963/300 = 719948 N/mm

Share of segment 2

F₂ = 40000·719948/824668 = 34921 N

Stress

σ₂ = 34921/1963 = 17.78 MPa (compression)

Answer

σ₂ = 17.78 MPa

3.Torsion of shafts

Answer: d = 46.78 mm

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Required polar moment

T = 200 N·m = 200000 N·mm

L = 0.8 m = 800 mm

G = 39 GPa = 39000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 200000·800/(39000·0.008727) = 470119 mm⁴

Diameter

d = (32J/π)^1/4 = 46.78 mm

Answer

d = 46.78 mm

In practice you would round up to the next standard size.

4.Torsion of shafts

Answer: d = 48.9 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 30000/157.1 = 191 N·m

Diameter for strength

T = 191 N·m = 190986 N·mm

d_τ = ∛(16·190986/(π·80)) = 22.99 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·190986/(π·39000·8.727 × 10⁻⁶))^1/4 = 48.9 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.99, 48.9) = 48.9 mm

Answer

d = 48.9 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: D (x = 3 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 = 28.33 kN, R₂ = 61.67 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 28.33, M_C = 56.67, M_D = 85 kN·m, M_E = 73.33, M_F = 61.67, M_G = 0

Answer

Mark D: M_D = 85 kN·m

6.Shear force & bending moment

Answer: M_max = 40.31 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 = 4·4 = 16 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 35·1.5 + 4·4·2

R_B = 84.5/4 = 21.13 kN

ΣF_y = 0: R_A = 51 − 21.13 = 29.88 kN

Where is V = 0?

V(1.5⁻) = R_A − w·1.5 = 23.88 kN

V(1.5⁺) = 23.88 − 35 = −11.13 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·1.5 − w·1.5²/2 = 29.88·1.5 − 4·1.5²/2 = 40.31 kN·m

Answer

M_max = 40.31 kN·m

It occurs at x = 1.5 m from A.

7.Centroid & second moment of area

Answer: ȳ = 25.08 mm

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 160·14 = 2240 mm², y₁ = 7 mm

two legs: A₂ = 2·14·66 = 1848 mm², y₂ = 14 + 66/2 = 47 mm

Centroid

ȳ = (2240·7 + 1848·47)/4088 = 25.08 mm

Answer

ȳ = 25.08 mm

8.Centroid & second moment of area

Answer: σ_j = 29.73 MPa

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 440 − 2·12 = 416 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (200·440³ − 190·416³)/12

I_x = 2.799 × 10⁸ mm⁴

Bending stress

y = 220 − 12 = 208 mm

σ = M·y/I = 4 × 10⁷·208/2.799 × 10⁸ = 29.73 MPa

Answer

σ_j = 29.73 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −58.28 °

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−60, 20 − 80) = −116.6°

Angle on the element

θ_p = −58.28°

Answer

θ_p = −58.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ₁ = 131.4 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·750000/(π·40³) = 119.4 MPa

τ = 16T/(πd³) = 16·500000/(π·40³) = 39.79 MPa

Mohr’s circle

C = σ/2 = 59.68 MPa

R = √(59.68² + 39.79²) = 71.73 MPa

σ₁ = C + R = 131.4 MPa

σ₂ = C − R = −12.05 MPa

Answer

σ₁ = 131.4 MPa

11.Beam deflection

Answer: δ_B = 53.03 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 = 17000 N, a = 500 mm, L = 2500 mm

EI = 9.35 × 10¹⁰ N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 7.576 mm

θ_P = Pa²/(2EI) = 0.02273 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 53.03 mm

Answer

δ_B = 53.03 mm

12.Beam deflection

Answer: I_req = 60 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 8000/360 = 22.22 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·5·8000⁴/(384·200000·22.22) = 6 × 10⁷ mm⁴

I_req = 60 × 10⁶ mm⁴

Answer

I_req = 60 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 81.88 kN

Euler buckling: a slender column becomes unstable at P_cr = 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.

I_min = h·b³/12 = 50·40³/12 = 266667 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·266667/1500² = 81881 N = 81.88 kN

Answer

P_cr = 81.88 kN

14.Column buckling

Answer: d = 57.08 mm

Required critical load

P_cr = n·P = 3·30000 = 90000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·2000²/(π³·70000))^(1/4) = 57.08 mm

Answer

d = 57.08 mm

Check: the slenderness KL/r is 140.2, so the column is slender and Euler applies.

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