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 aluminium bar under the axial load.

P = 50 kNd = 20 mmL = 2.5 mE = 70 GPa

δ = mm

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 = 65 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 400 N·md = 30 mmL = 1.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft transmits 50 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 = 50 kWN = 1500 rpmτ_allow = 80 MPaθ_allow = 0.5°/mG = 80 GPa

d = mm

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 = 5 mP = 20 kN at 2 mw = 4 kN/m from 0 to 5 mx = 1 m

M_x = kN·m

6.Shear force & bending momentHard

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

L = 4 mP = 45 kN at 1.5 mw = 12 kN/m from 0 to 4 mx = 1.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 170 mm and inside diameter d = 138 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 170 mmd = 138 mm

r = mm

8.Centroid & second moment of areaHard

A 150 mm × 300 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 90 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 300 mmcut-out: 50 × 20 mmhole centre: y = 90 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −40 MPa, σᵧ = 100 MPa and τₓᵧ = 30 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 = −40 MPaσ_y = 100 MPaτ_xy = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 110 MPa, σᵧ = −80 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 110 MPaσ_y = −80 MPaτ_xy = 40 MPaθ = -45°

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 = 2 kN/mL = 2 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)b × h = 30 × 60 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 5.684 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·20²/4 = 314.2 mm²

σ = P/A = 50·1000/314.2 = 159.2 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 159.2/70000 = 0.002274 = 2274 µε

Elongation

δ = ε·L = 0.002274·2500 mm = 5.684 mm

Answer

δ = 5.684 mm

2.Axial stress & strain

Answer: σ_max = 574.7 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 = 51.73 MPa

σ₂ = 574.7 MPa

Largest stress

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

σ_max = 574.7 MPa

Answer

σ_max = 574.7 MPa

3.Torsion of shafts

Answer: φ = 16.63 °

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.

Polar second moment of area

J = π·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1500/(26000·79522) = 0.2902 rad

φ = 0.2902·180/π = 16.63°

Answer

φ = 16.63°

4.Torsion of shafts

Answer: d = 46.42 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 50000/157.1 = 318.3 N·m

Diameter for strength

T = 318.3 N·m = 318310 N·mm

d_τ = ∛(16·318310/(π·80)) = 27.26 mm

Diameter for stiffness

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

G = 80 GPa = 80000 MPa

d_φ = (32·318310/(π·80000·8.727 × 10⁻⁶))^1/4 = 46.42 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(27.26, 46.42) = 46.42 mm

Answer

d = 46.42 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = −52 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: −4·4²/2 = −32 kN·m

M_x = −20 − 32 = −52 kN·m

Answer

M_x = −52 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_x = 64.69 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·4 = 48 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 45·1.5 + 12·4·2

R_B = 163.5/4 = 40.88 kN

ΣF_y = 0: R_A = 93 − 40.88 = 52.13 kN

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.

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

M_x = 52.13·1.5 − 12·1.5·0.75 = 64.69 kN·m

Answer

M_x = 64.69 kN·m

7.Centroid & second moment of area

Answer: r = 54.74 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 7741 mm²

I = π(D⁴ − d⁴)/64 = 2.32 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(2.32 × 10⁷/7741) = 54.74 mm

Answer

r = 54.74 mm

8.Centroid & second moment of area

Answer: I_x = 333.8 × 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 = (45000·150 − 1000·90)/(45000 − 1000) = 151.4 mm

Full plate about ȳ

150·300³/12 + 45000·−1.364² = 337.6 × 10⁶ mm⁴

Hole about ȳ

50·20³/12 + 1000·−61.36² = 3.799 × 10⁶ mm⁴

Subtract

I_x = 333.8 × 10⁶ mm⁴

Answer

I_x = 333.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 78.4 °

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(2·30, −40 − 100) = 156.8°

Angle on the element

θ_p = 78.4°

Answer

θ_p = 78.4°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 95 MPa

Transformation equation

τ_θ = −95·−1 + 40·0 = 95 MPa

Answer

τ_θ = 95 MPa

11.Beam deflection

Answer: δ = 3.03 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.

w = 2 N/mm, L = 2000 mm

EI = 1.32 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 2·2000⁴/(8·1.32 × 10¹²) = 3.03 mm

Answer

δ = 3.03 mm

12.Beam deflection

Answer: h_req = 223.1 mm

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 = 2.5 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 5000/250 = 20 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·5000⁴/(384·11000·20) = 9.248 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·9.248 × 10⁷/100)^1/3 = 223.1 mm

Answer

h_req = 223.1 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 29.61 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·135000/3000² = 29609 N = 29.61 kN

Answer

P_cr = 29.61 kN

14.Column buckling

Answer: σ_cr = 7.896 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·5000 = 5000 mm

KL/r = 5000/10 = 500

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/500² = 7.896 MPa

Answer

σ_cr = 7.896 MPa

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