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 axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 70 kNd = 16 mmL = 3 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 115 kNsegment 1: Ø40 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 750 N·md = 25 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 1.2 m from A and 3 m from B. Find the maximum shear stress τ_max in the shaft.

d = 50 mmT₀ = 400 N·ma = 1.2 mb = 3 m

τ_max = MPa

5.Shear force & bending momentMedium

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 M_x 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 = 30 kN at 2 mw = 12 kN/m from 0 to 2 mx = 1 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 35 kN at 4 mw = 5 kN/m from 0 to 5 mx = 3.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 150 mm × 240 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 156 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 240 mmd = 30 mmhole centre: y = 156 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 170 mm wide and 22 mm thick, an overall depth of 280 mm and a web 10 mm thick. It carries a bending moment of 150 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 = 170 mmH = 280 mmt_f = 22 mmt_w = 10 mmM = 150 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 90 MPa, σᵧ = 120 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 90 MPaσ_y = 120 MPaτ_xy = 60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 120 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 120 MPaσ_y = 110 MPaτ_xy = −20 MPaθ = 20°

τ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 5 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 5 mw = 6 kN/m

R_B = kN

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 = 4 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

14.Column bucklingHard

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

L = 6 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 2976 µε

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·16²/4 = 201.1 mm²

σ = P/A = 70·1000/201.1 = 348.2 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 348.2/117000 = 0.002976 = 2976 µε

Answer

ε = 2976 µε

2.Axial stress & strain

Answer: σ_max = 1017 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 91.51 MPa

σ₂ = 1017 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 1017 MPa

Answer

σ_max = 1017 MPa

3.Torsion of shafts

Answer: φ = 22.99 °

Polar second moment of area

J = π·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·800/(39000·38350) = 0.4012 rad

φ = 0.4012·180/π = 22.99°

Answer

φ = 22.99°

4.Torsion of shafts

Answer: τ_max = 11.64 MPa

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 400

T_A·a = T_B·b

T_A = 400·3/4.2 = 285.7 N·m

T_B = 400·1.2/4.2 = 114.3 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τ_max = 16·285714/(π·50³) = 11.64 MPa

Answer

τ_max = 11.64 MPa

5.Shear force & bending moment

Answer: M_x = −36 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: −30·1 = −30 kN·m

w: −12·1²/2 = −6 kN·m

M_x = −30 − 6 = −36 kN·m

Answer

M_x = −36 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_x = 37.63 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·5 = 25 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 35·4 + 5·5·2.5

R_B = 202.5/5 = 40.5 kN

ΣF_y = 0: R_A = 60 − 40.5 = 19.5 kN

Cut at x = 3.5 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 19.5·3.5 − 5·3.5·1.75 = 37.63 kN·m

Answer

M_x = 37.63 kN·m

7.Centroid & second moment of area

Answer: I_x = 171.8 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (36000·120 − 706.9·156)/(36000 − 706.9) = 119.3 mm

Full plate about ȳ

150·240³/12 + 36000·0.721² = 172.8 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·36.72² = 0.9929 × 10⁶ mm⁴

Subtract

I_x = 171.8 × 10⁶ mm⁴

Answer

I_x = 171.8 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_j = 130.4 MPa

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.

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 = 280 − 2·22 = 236 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (170·280³ − 160·236³)/12

I_x = 1.357 × 10⁸ mm⁴

Bending stress

y = 140 − 22 = 118 mm

σ = M·y/I = 1.5 × 10⁸·118/1.357 × 10⁸ = 130.4 MPa

Answer

σ_j = 130.4 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 61.85 MPa

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.

Radius of Mohr’s circle

τ_max = R = √(((90 − 120)/2)² + 60²) = 61.85 MPa

Answer

τ_max = 61.85 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = −18.53 MPa

Transformation equation

τ_θ = −5·0.6428 + −20·0.766 = −18.53 MPa

Answer

τ_θ = −18.53 MPa

11.Beam deflection

Answer: δ = 4.152 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 = 5 N/mm, L = 2500 mm

EI = 5.88 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 5·2500⁴/(8·5.88 × 10¹²) = 4.152 mm

Answer

δ = 4.152 mm

12.Beam deflection

Answer: R_B = 11.25 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·6·5/8 = 11.25 kN

Answer

R_B = 11.25 kN

13.Column buckling

Answer: P_cr = 3.312 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 round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·19175/2000² = 3312 N = 3.312 kN

Answer

P_cr = 3.312 kN

14.Column buckling

Answer: σ_cr = 5.483 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 2·6000 = 12000 mm

KL/r = 12000/20 = 600

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σ_cr = 5.483 MPa

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