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 aluminium bar, in microstrain (µε = 10⁻⁶).

P = 90 kNd = 30 mmL = 1.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 60 kNsegment 1: Ø35 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 0.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

T = 2000 N·md_i/d_o = 0.5τ_allow = 60 MPa

d_o = mm

4.Torsion of shaftsHard

A solid brass shaft transmits 15 kW at 1000 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 15 kWN = 1000 rpmτ_allow = 60 MPaθ_allow = 0.5°/mG = 39 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 3 m from A. Find the bending moment M_x at the section x = 1.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 3 mP = 40 kN at 3 mw = 12 kN/m from 0 to 3 mx = 1.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 40 kN at 2.5 mw = 12 kN/m from 0 to 4 mx = 3 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 144 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 240 mmd = 50 mmhole centre: y = 144 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 60 mm wide and 50 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 300 mmcut-out: 60 × 50 mmhole centre: y = 120 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −40 MPa, σᵧ = 0 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = −40 MPaσ_y = 0 MPaτ_xy = −20 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −50 MPa, σᵧ = 60 MPa and τₓᵧ = 50 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 = −50 MPaσ_y = 60 MPaτ_xy = 50 MPaθ = 50°

τ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2 m from A. Find the deflection δ_B at the free end B.

P = 23 kNa = 2 mL = 2.5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δ_max.

P = 26 kNL = 5 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 2.5 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

14.Column bucklingHard

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

L = 2.5 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1819 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·30²/4 = 706.9 mm²

σ = P/A = 90·1000/706.9 = 127.3 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 127.3/70000 = 0.001819 = 1819 µε

Answer

ε = 1819 µε

2.Axial stress & strain

Answer: δ = 1.44 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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 62.36 MPa

σ₂ = 191 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 62.36·2000/200000 = 0.6236 mm

δ₂ = σ₂L₂/E₂ = 191·500/117000 = 0.8162 mm

δ = δ₁ + δ₂ = 1.44 mm

Answer

δ = 1.44 mm

3.Torsion of shafts

Answer: d_o = 56.58 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.

Outer diameter from the allowable stress

T = 2000 N·m = 2 × 10⁶ N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·60·0.9375)) = 56.58 mm

d_i = 0.5·56.58 = 28.29 mm

Answer

d_o = 56.58 mm

A solid shaft would need d = 55.37 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 45.5 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 15000/104.7 = 143.2 N·m

Diameter for strength

T = 143.2 N·m = 143239 N·mm

d_τ = ∛(16·143239/(π·60)) = 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 = (32·143239/(π·39000·8.727 × 10⁻⁶))^1/4 = 45.5 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.99, 45.5) = 45.5 mm

Answer

d = 45.5 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = −73.5 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: −40·1.5 = −60 kN·m

w: −12·1.5²/2 = −13.5 kN·m

M_x = −60 − 13.5 = −73.5 kN·m

Answer

M_x = −73.5 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_x = 43 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 = 40·2.5 + 12·4·2

R_B = 196/4 = 49 kN

ΣF_y = 0: R_A = 88 − 49 = 39 kN

Cut at x = 3 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 = 39·3 − 40·(3 − 2.5) − 12·3·1.5 = 43 kN·m

Answer

M_x = 43 kN·m

7.Centroid & second moment of area

Answer: I_x = 113.7 × 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 = (24000·120 − 1963·144)/(24000 − 1963) = 117.9 mm

Full plate about ȳ

100·240³/12 + 24000·2.138² = 115.3 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·26.14² = 1.648 × 10⁶ mm⁴

Subtract

I_x = 113.7 × 10⁶ mm⁴

Answer

I_x = 113.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 266.4 × 10⁶ mm⁴

Centroid

ȳ = (36000·150 − 3000·120)/(36000 − 3000) = 152.7 mm

Full plate about ȳ

120·300³/12 + 36000·−2.727² = 270.3 × 10⁶ mm⁴

Hole about ȳ

60·50³/12 + 3000·−32.73² = 3.838 × 10⁶ mm⁴

Subtract

I_x = 266.4 × 10⁶ mm⁴

Answer

I_x = 266.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = 67.5 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −20/28.28 = −0.7071, sin 2θ_s = −(−20)/28.28 = 0.7071

2θ_s = atan2(20, −20) = 135°

Angle on the element

θ_s = 67.5°

check: θ_s − θ_p = 67.5 − (−67.5) = 135°

Answer

θ_s = 67.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 45.48 MPa

Transformation equation

τ_θ = −−55·0.9848 + 50·−0.1736 = 45.48 MPa

Answer

τ_θ = 45.48 MPa

11.Beam deflection

Answer: δ_B = 49.61 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 = 23000 N, a = 2000 mm, L = 2500 mm

EI = 1.7 × 10¹² 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) = 36.08 mm

θ_P = Pa²/(2EI) = 0.02706 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 49.61 mm

Answer

δ_B = 49.61 mm

12.Beam deflection

Answer: δ_max = 45.78 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 = 26000 N, L = 5000 mm

EI = 2.52 × 10¹² N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 1667 mm

δ₁ = 22.89 mm

Add both loads

δ_max = 2·δ₁ = 23PL³/(648EI) = 45.78 mm

Answer

δ_max = 45.78 mm

13.Column buckling

Answer: P_cr = 4.326 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 = 0.7·2500 = 1750 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·19175/1750² = 4326 N = 4.326 kN

Answer

P_cr = 4.326 kN

14.Column buckling

Answer: σ_cr = 31.58 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·2500 = 1250 mm

KL/r = 1250/5 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa

Answer

σ_cr = 31.58 MPa

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