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 = 15 kNd = 20 mmL = 0.5 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 total elongation δ.

P = 65 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 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 = 50 mmL = 2 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 500 N·m is applied at C, 1 m from A and 0.8 m from B. Find the angle of rotation φ_C of section C, in degrees.

d = 30 mmT₀ = 500 N·ma = 1 mb = 0.8 mG = 39 GPa

φ_C = °

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 5 kN/m from 0 to 10 mx = 4 m

M_x = kN·m

6.Shear force & bending momentHard

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

55 kN @ 0 m45 kN @ 6 m4 kN/m, 1–5 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 170 mm wide and 24 mm thick, an overall depth of 400 mm and a web 14 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 170 mmH = 400 mmt_f = 24 mmt_w = 14 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 160 mm wide and 14 mm thick, an overall depth of 440 mm and a web 10 mm thick. It carries a bending moment of 30 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 = 160 mmH = 440 mmt_f = 14 mmt_w = 10 mmM = 30 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 60 MPa, σᵧ = 50 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 60 MPaσ_y = 50 MPaτ_xy = −30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 200 N·m together with a torque of 1500 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 40 mmM = 200 N·mT = 1500 N·m

τ_max = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 15 kN/mL = 2 mE = 70 GPa (aluminium)I = 8.5 × 10⁶ mm⁴

θ_A = rad

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 = 24 kNw = 6 kN/mL = 1.5 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 1.5 mE = 200 GPa (steel)d = 30 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 408.1 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·20²/4 = 314.2 mm²

σ = P/A = 15·1000/314.2 = 47.75 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 47.75/117000 = 0.0004081 = 408.1 µε

Answer

ε = 408.1 µε

2.Axial stress & strain

Answer: δ = 3.539 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·40²/4 = 1257 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 51.73 MPa

σ₂ = 323.3 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 51.73·1500/100000 = 0.7759 mm

δ₂ = σ₂L₂/E₂ = 323.3·1000/117000 = 2.763 mm

δ = δ₁ + δ₂ = 3.539 mm

Answer

δ = 3.539 mm

3.Torsion of shafts

Answer: φ = 0.9577 °

Polar second moment of area

J = π·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2000/(39000·613592) = 0.01672 rad

φ = 0.01672·180/π = 0.9577°

Answer

φ = 0.9577°

4.Torsion of shafts

Answer: φ_C = 4.105 °

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 500

T_A·a = T_B·b

T_A = 500·0.8/1.8 = 222.2 N·m

T_B = 500·1/1.8 = 277.8 N·m

Rotation of C

Use segment AC: it is fixed at A and carries T_A.

J = π·30⁴/32 = 79522 mm⁴

G = 39 GPa = 39000 MPa

φ_C = T_A·a/(GJ) = 222222·1000/(39000·79522) = 0.07165 rad = 4.105°

Answer

φ_C = 4.105°

5.Shear force & bending moment

Answer: M_x = 60 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 5·10·5

R_B = 250/10 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 4 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 = 25·4 − 5·4·2 = 60 kN·m

Answer

M_x = 60 kN·m

6.Shear force & bending moment

Answer: B (x = 1 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 65.5 kN, R₂ = 50.5 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = −55 kN·m, M_C = −46.5, M_D = −42, M_E = −41.5, M_F = −45, M_G = 0

Answer

Mark B: M_B = −55 kN·m

7.Centroid & second moment of area

Answer: I_x = 339.7 × 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 = 400 − 2·24 = 352 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (170·400³ − 156·352³)/12

I_x = 3.397 × 10⁸ mm⁴ = 339.7 × 10⁶ mm⁴

Answer

I_x = 339.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_j = 23.62 MPa

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.

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 − 2·14 = 412 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (160·440³ − 150·412³)/12

I_x = 2.616 × 10⁸ mm⁴

Bending stress

y = 220 − 14 = 206 mm

σ = M·y/I = 3 × 10⁷·206/2.616 × 10⁸ = 23.62 MPa

Answer

σ_j = 23.62 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 30.41 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 = √(((60 − 50)/2)² + −30²) = 30.41 MPa

Answer

τ_max = 30.41 MPa

10.Mohr’s circle & stress transformation

Answer: τ_max = 120.4 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·200000/(π·40³) = 31.83 MPa

τ = 16T/(πd³) = 16·1500000/(π·40³) = 119.4 MPa

Mohr’s circle

C = σ/2 = 15.92 MPa

R = √(15.92² + 119.4²) = 120.4 MPa

Answer

τ_max = 120.4 MPa

11.Beam deflection

Answer: θ_A = 0.008403 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 15 N/mm, L = 2000 mm

EI = 5.95 × 10¹¹ N·mm²

End slope

θ_A = wL³/(24EI) = 15·2000³/(24·5.95 × 10¹¹) = 0.008403 rad

Answer

θ_A = 0.008403 rad

That is 0.4815°.

12.Beam deflection

Answer: δ_max = 5.26 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 = 24000 N, w = 6 N/mm, L = 1500 mm, EI = 3.96 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 24000·1500³/(48·3.96 × 10¹¹) = 4.261 mm

δ_w = 5wL⁴/(384EI) = 5·6·1500⁴/(384·3.96 × 10¹¹) = 0.9988 mm

Add them

δ_max = 4.261 + 0.9988 = 5.26 mm

Answer

δ_max = 5.26 mm

13.Column buckling

Answer: P_cr = 139.5 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·1500 = 750 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/750² = 139528 N = 139.5 kN

Answer

P_cr = 139.5 kN

14.Column buckling

Answer: σ_cr = 15.67 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 = 0.7·3000 = 2100 mm

KL/r = 2100/10 = 210

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/210² = 15.67 MPa

Answer

σ_cr = 15.67 MPa

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