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 = 65 kNd = 12 mmL = 2 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 = 75 kNsegment 1: Ø25 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 0.5 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 50 mm and length 0.5 m; segment BC has diameter 40 mm and length 1.5 m. A torque of 250 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 250 N·md₁ = 50 mm, L₁ = 0.5 md₂ = 40 mm, L₂ = 1.5 mG = 80 GPa

φ_C = °

4.Torsion of shaftsHard

A solid aluminium shaft transmits 30 kW at 750 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 = 750 rpmτ_allow = 80 MPaθ_allow = 0.5°/mG = 26 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 4 kN/m from 0 to 5 mx = 2.5 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment M_B at support B. Use sagging moments positive and upward shear on the left face positive.

L = 6.5 mw = 6 kN/m from 0 to 6.5 msupports at x = 0 and x = 5 m

M_B = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 240 mm wide with two upright legs, 150 mm high overall. Base and legs are all 18 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 240 mmh = 150 mmt = 18 mm

ȳ = mm

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 40 MPa in tension and 90 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 280 mmt_f = 10 mmt_w = 18 mmh_w = 120 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = 20 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 50 MPaσ_y = 20 MPaτ_xy = 60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 0 MPa, σᵧ = 40 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 0 MPaσ_y = 40 MPaτ_xy = −50 MPaθ = 45°

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 = 4 kN/mL = 4 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 22 kNw = 15 kN/mL = 3 mE = 200 GPa (steel)I = 23 × 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 = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 5 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 4912 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·12²/4 = 113.1 mm²

σ = P/A = 65·1000/113.1 = 574.7 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 574.7/117000 = 0.004912 = 4912 µε

Answer

ε = 4912 µε

2.Axial stress & strain

Answer: σ_max = 373 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 152.8 MPa

σ₂ = 373 MPa

Largest stress

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

σ_max = 373 MPa

Answer

σ_max = 373 MPa

3.Torsion of shafts

Answer: φ_C = 1.215 °

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 moments

J₁ = π·50⁴/32 = 613592 mm⁴

J₂ = π·40⁴/32 = 251327 mm⁴

Twist of each segment

T = 250 N·m = 250000 N·mm

G = 80 GPa = 80000 MPa

φ_AB = 250000·500/(80000·613592) = 0.002546 rad

φ_BC = 250000·1500/(80000·251327) = 0.01865 rad

Total twist

φ_C = 0.002546 + 0.01865 = 0.0212 rad = 1.215°

Answer

φ_C = 1.215°

4.Torsion of shafts

Answer: d = 64.35 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 30000/78.54 = 382 N·m

Diameter for strength

T = 382 N·m = 381972 N·mm

d_τ = ∛(16·381972/(π·80)) = 28.97 mm

Diameter for stiffness

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

G = 26 GPa = 26000 MPa

d_φ = (32·381972/(π·26000·8.727 × 10⁻⁶))^1/4 = 64.35 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(28.97, 64.35) = 64.35 mm

Answer

d = 64.35 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 12.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 4·5 = 20 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 4·5·2.5

R_B = 50/5 = 10 kN

ΣF_y = 0: R_A = 20 − 10 = 10 kN

Cut at x = 2.5 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 = 10·2.5 − 4·2.5·1.25 = 12.5 kN·m

Answer

M_x = 12.5 kN·m

6.Shear force & bending moment

Answer: M_B = −6.75 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 6·6.5 = 39 kN, acting at x̄ = 3.25 m

ΣM_A = 0: R_B·5 = 6·6.5·3.25

R_B = 126.8/5 = 25.35 kN

ΣF_y = 0: R_A = 39 − 25.35 = 13.65 kN

Cut at B and keep the overhang

The load on the overhang, w·a, acts at a/2 from B and makes the beam hog over the support.

M_B = −w·a²/2 = −6·1.5²/2 = −6.75 kN·m

Answer

M_B = −6.75 kN·m

7.Centroid & second moment of area

Answer: ȳ = 48.29 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 = 240·18 = 4320 mm², y₁ = 9 mm

two legs: A₂ = 2·18·132 = 4752 mm², y₂ = 18 + 132/2 = 84 mm

Centroid

ȳ = (4320·9 + 4752·84)/9072 = 48.29 mm

Answer

ȳ = 48.29 mm

8.Centroid & second moment of area

Answer: M_allow = 3.213 kN·m

Centroid

ȳ = (2800·125 + 2160·60)/4960 = 96.69 mm

top edge: c = 130 − 96.69 = 33.31 mm

Second moment of area

I = 280·10³/12 + 2800·28.31² + 18·120³/12 + 2160·36.69² = 7.767 × 10⁶ mm⁴

Moment allowed by each edge

Sagging: the top is in compression, the bottom in tension.

top edge: M = 90·7.767 × 10⁶/33.31 = 20.99 kN·m

bottom edge: M = 40·7.767 × 10⁶/96.69 = 3.213 kN·m

Governing value

M_allow = min = 3.213 kN·m (bottom edge)

Answer

M_allow = 3.213 kN·m

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 = √(((50 − 20)/2)² + 60²) = 61.85 MPa

Answer

τ_max = 61.85 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 20 MPa

Transformation equation

τ_θ = −−20·1 + −50·0 = 20 MPa

Answer

τ_θ = 20 MPa

11.Beam deflection

Answer: δ_max = 7.576 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 = 4 N/mm, L = 4000 mm

EI = 1.76 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·4000⁴/(384·1.76 × 10¹²) = 7.576 mm

Answer

δ_max = 7.576 mm

12.Beam deflection

Answer: δ_max = 6.129 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 = 22000 N, w = 15 N/mm, L = 3000 mm, EI = 4.6 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 22000·3000³/(48·4.6 × 10¹²) = 2.69 mm

δ_w = 5wL⁴/(384EI) = 5·15·3000⁴/(384·4.6 × 10¹²) = 3.439 mm

Add them

δ_max = 2.69 + 3.439 = 6.129 mm

Answer

δ_max = 6.129 mm

13.Column buckling

Answer: P_cr = 177.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2800² = 177179 N = 177.2 kN

Answer

P_cr = 177.2 kN

14.Column buckling

Answer: σ_cr = 11.05 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 = 1·5000 = 5000 mm

KL/r = 5000/20 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/250² = 11.05 MPa

Answer

σ_cr = 11.05 MPa

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