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

P = 45 kNd = 16 mmL = 2.5 mE = 100 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 = 120 kNsegment 1: Ø30 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1.5 m, copper (E = 117 GPa)

δ = mm

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 = 200 N·md = 30 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 20 kW at 300 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 300 rpmτ_allow = 75 MPaθ_allow = 0.25°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 6 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 4 kN/m from 0 to 8 mx = 6 m

V_x = kN

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 8 mw = 4 kN/m from 0 to 8 msupports at x = 0 and x = 6 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 280 mm wide with two upright legs, 120 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 280 mmh = 120 mmt = 8 mm

ȳ = mm

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 110 mm wide and 20 mm thick, an overall depth of 440 mm and a web 10 mm thick. It carries a bending moment of 40 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 = 110 mmH = 440 mmt_f = 20 mmt_w = 10 mmM = 40 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = −60 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 = 50 MPaσ_y = −60 MPaτ_xy = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −50 MPa, σᵧ = 100 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σ_x = −50 MPaσ_y = 100 MPaτ_xy = −50 MPaθ = 35°

σ_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 = 5 kN/mL = 4 mE = 70 GPa (aluminium)I = 8.5 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 2 kN/mL = 5 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 30 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 30 kNn = 2L = 2.5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 2238 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 = 45·1000/201.1 = 223.8 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 223.8/100000 = 0.002238 = 2238 µε

Answer

ε = 2238 µε

2.Axial stress & strain

Answer: δ = 8.535 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·20²/4 = 314.2 mm²

σ₁ = 169.8 MPa

σ₂ = 382 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 169.8·1500/70000 = 3.638 mm

δ₂ = σ₂L₂/E₂ = 382·1500/117000 = 4.897 mm

δ = δ₁ + δ₂ = 8.535 mm

Answer

δ = 8.535 mm

3.Torsion of shafts

Answer: φ = 3.695 °

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) = 200·1000·1000/(39000·79522) = 0.06449 rad

φ = 0.06449·180/π = 3.695°

Answer

φ = 3.695°

4.Torsion of shafts

Answer: d = 78.57 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 20000/31.42 = 636.6 N·m

Diameter for strength

T = 636.6 N·m = 636620 N·mm

d_τ = ∛(16·636620/(π·75)) = 35.1 mm

Diameter for stiffness

θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·636620/(π·39000·4.363 × 10⁻⁶))^1/4 = 78.57 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(35.1, 78.57) = 78.57 mm

Answer

d = 78.57 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: V_x = −8 kN

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 4·8 = 32 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 4·8·4

R_B = 128/8 = 16 kN

ΣF_y = 0: R_A = 32 − 16 = 16 kN

Cut at x = 6 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.

V = R_A − (all downward loads left of the cut)

V_x = 16 − 4·6 = −8 kN

Answer

V_x = −8 kN

6.Shear force & bending moment

Answer: M_max = 14.22 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 = 4·8 = 32 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·6 = 4·8·4

R_B = 128/6 = 21.33 kN

ΣF_y = 0: R_A = 32 − 21.33 = 10.67 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 10.67/4 = 2.667 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 10.67²/(2·4) = 14.22 kN·m

Answer

M_max = 14.22 kN·m

7.Centroid & second moment of area

Answer: ȳ = 30.67 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 = 280·8 = 2240 mm², y₁ = 4 mm

two legs: A₂ = 2·8·112 = 1792 mm², y₂ = 8 + 112/2 = 64 mm

Centroid

ȳ = (2240·4 + 1792·64)/4032 = 30.67 mm

Answer

ȳ = 30.67 mm

8.Centroid & second moment of area

Answer: σ_j = 32.32 MPa

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 = 440 − 2·20 = 400 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (110·440³ − 100·400³)/12

I_x = 2.475 × 10⁸ mm⁴

Bending stress

y = 220 − 20 = 200 mm

σ = M·y/I = 4 × 10⁷·200/2.475 × 10⁸ = 32.32 MPa

Answer

σ_j = 32.32 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 14.31 °

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, 50 − −60) = 28.61°

Angle on the element

θ_p = 14.31°

Answer

θ_p = 14.31°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = −47.64 MPa

Transformation equation

σ_θ = 25 + −75·0.342 + −50·0.9397 = −47.64 MPa

Answer

σ_θ = −47.64 MPa

11.Beam deflection

Answer: δ_max = 28.01 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 = 4000 mm

EI = 5.95 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·5·4000⁴/(384·5.95 × 10¹¹) = 28.01 mm

Answer

δ_max = 28.01 mm

12.Beam deflection

Answer: I_req = 5.859 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·2·5000⁴/(384·200000·13.89) = 5.859 × 10⁶ mm⁴

I_req = 5.859 × 10⁶ mm⁴

Answer

I_req = 5.859 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 9.462 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 = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/8000² = 9462 N = 9.462 kN

Answer

P_cr = 9.462 kN

14.Column buckling

Answer: d = 81.55 mm

Required critical load

P_cr = n·P = 2·30000 = 60000 N

KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·60000·5000²/(π³·70000))^(1/4) = 81.55 mm

Answer

d = 81.55 mm

Check: the slenderness KL/r is 245.2, so the column is slender and Euler applies.

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