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 = 105 kNd = 12 mmL = 1.5 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 total elongation δ.

P = 55 kNsegment 1: Ø30 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel 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 = 25 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 50 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 50 kWN = 300 rpmd_o = 80 mmd_i = 48 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 25 kN·m at x = 2 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 5 mM₀ = 25 kN·m clockwise, at 2 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 55 kN at 1 mw = 6 kN/m from 0 to 5 mx = 1 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 160 mm wide and 22 mm thick, an overall depth of 260 mm and a web 6 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 160 mmH = 260 mmt_f = 22 mmt_w = 6 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 60 mm wide and 20 mm high, centred on the 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 = 120 mmh = 240 mmcut-out: 60 × 20 mmhole centre: y = 156 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = 40 MPa and τₓᵧ = −30 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 70 MPaσ_y = 40 MPaτ_xy = −30 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 120 MPa, σᵧ = −30 MPa and τₓᵧ = −30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 120 MPaσ_y = −30 MPaτ_xy = −30 MPaθ = 50°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 8 kN/mL = 1.5 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2 m from support B. Find the maximum deflection δ_max of the beam.

P = 20 kNb = 2 mL = 6 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ_max = 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 = 2 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 80 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 80 kNn = 2L = 6 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 19.89 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·12²/4 = 113.1 mm²

σ = P/A = 105·1000/113.1 = 928.4 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 928.4/70000 = 0.01326 = 13263 µε

Elongation

δ = ε·L = 0.01326·1500 mm = 19.89 mm

Answer

δ = 19.89 mm

2.Axial stress & strain

Answer: δ = 5.075 mm

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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 77.81 MPa

σ₂ = 273.5 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 77.81·1500/100000 = 1.167 mm

δ₂ = σ₂L₂/E₂ = 273.5·1000/70000 = 3.908 mm

δ = δ₁ + δ₂ = 5.075 mm

Answer

δ = 5.075 mm

3.Torsion of shafts

Answer: φ = 14.94 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·2000/(80000·38350) = 0.2608 rad

φ = 0.2608·180/π = 14.94°

Answer

φ = 14.94°

4.Torsion of shafts

Answer: τ_max = 18.19 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 50000/31.42 = 1592 N·m

Polar second moment of area

J = π(80⁴ − 48⁴)/32 = 3.5 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 1.592 × 10⁶·40/3.5 × 10⁶ = 18.19 MPa

Answer

τ_max = 18.19 MPa

5.Shear force & bending moment

Answer: M_x = −10 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 25/5 = 5 kN (upward)

R_A = −R_B = −5 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −5·2 = −10 kN·m

Answer

M_x = −10 kN·m

6.Shear force & bending moment

Answer: M_x = 56 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·5 = 30 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 55·1 + 6·5·2.5

R_B = 130/5 = 26 kN

ΣF_y = 0: R_A = 85 − 26 = 59 kN

Cut at x = 1 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 = 59·1 − 6·1·0.5 = 56 kN·m

Answer

M_x = 56 kN·m

7.Centroid & second moment of area

Answer: I_x = 105 × 10⁶ mm⁴

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.

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 = 260 − 2·22 = 216 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (160·260³ − 154·216³)/12

I_x = 1.05 × 10⁸ mm⁴ = 105 × 10⁶ mm⁴

Answer

I_x = 105 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 136.6 × 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 = (28800·120 − 1200·156)/(28800 − 1200) = 118.4 mm

Full plate about ȳ

120·240³/12 + 28800·1.565² = 138.3 × 10⁶ mm⁴

Hole about ȳ

60·20³/12 + 1200·37.57² = 1.733 × 10⁶ mm⁴

Subtract

I_x = 136.6 × 10⁶ mm⁴

Answer

I_x = 136.6 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 44.27 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 55 MPa, R = √(15² + (−30)²) = 33.54 MPa

σ₁ = 88.54 MPa, σ₂ = 21.46 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(67.08, 88.54, 21.46)/2 = 44.27 MPa

Answer

τ_abs = 44.27 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: τ_θ = −68.65 MPa

Transformation equation

τ_θ = −75·0.9848 + −30·−0.1736 = −68.65 MPa

Answer

τ_θ = −68.65 MPa

11.Beam deflection

Answer: δ_max = 1.332 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 = 8 N/mm, L = 1500 mm

EI = 3.96 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·1500⁴/(384·3.96 × 10¹¹) = 1.332 mm

Answer

δ_max = 1.332 mm

12.Beam deflection

Answer: δ_max = 9.216 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.

P = 20000 N, b = 2000 mm, L = 6000 mm

EI = 8.4 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((6000² − 2000²)/3) = 3266 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 9.216 mm

Answer

δ_max = 9.216 mm

13.Column buckling

Answer: P_cr = 151.4 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/2000² = 151398 N = 151.4 kN

Answer

P_cr = 151.4 kN

14.Column buckling

Answer: d = 73.46 mm

Required critical load

P_cr = n·P = 2·80000 = 160000 N

KL = 0.7·6000 = 4200 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·160000·4200²/(π³·200000))^(1/4) = 73.46 mm

Answer

d = 73.46 mm

Check: the slenderness KL/r is 228.7, so the column is slender and Euler applies.

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