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 brass bar under the axial load.

P = 70 kNd = 12 mmL = 0.5 mE = 100 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 = 120 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 0.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 10 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 250 mm rotate 9.12° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 10 mmL = 250 mmφ = 9.12°

G = GPa

4.Torsion of shaftsHard

A solid brass shaft transmits 20 kW at 1450 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 1450 rpmτ_allow = 60 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

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 @ 8 m10 kN @ 2 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 7.5 mw = 10 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m

x_c = m

7.Centroid & second moment of areaMedium

A hollow rectangular (box) section has outer width B = 100 mm, outer height H = 175 mm and a uniform wall thickness t = 6 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 100 mmH = 175 mmt = 6 mm

r = mm

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 80 mm wide and 40 mm high, centred on the vertical centre line with its centre 90 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: 80 × 40 mmhole centre: y = 90 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa and σᵧ = 70 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 135 MPa?

σ_x = 100 MPaσ_y = 70 MPaσ₁ = 135 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 130 MPa, σᵧ = −60 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 130 MPaσ_y = −60 MPaτ_xy = −60 MPaθ = 35°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 28 kNa = 1 mL = 8 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_P = 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 = 6 kN/mL = 3 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

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 = 2 mE = 200 GPa (steel)b × h = 20 × 40 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 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 85 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 3.095 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 = 70·1000/113.1 = 618.9 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 618.9/100000 = 0.006189 = 6189 µε

Elongation

δ = ε·L = 0.006189·500 mm = 3.095 mm

Answer

δ = 3.095 mm

2.Axial stress & strain

Answer: δ = 3.41 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·20²/4 = 314.2 mm²

σ₁ = 95.49 MPa

σ₂ = 382 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 95.49·500/70000 = 0.6821 mm

δ₂ = σ₂L₂/E₂ = 382·500/70000 = 2.728 mm

δ = δ₁ + δ₂ = 3.41 mm

Answer

δ = 3.41 mm

3.Torsion of shafts

Answer: G = 79.99 GPa

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 moment and angle

J = π·10⁴/32 = 981.7 mm⁴

φ = 9.12·π/180 = 0.1592 rad

Shear modulus

T = 50 N·m = 50000 N·mm

G = TL/(Jφ) = 50000·250/(981.7·0.1592) = 79990 MPa

G = 79.99 GPa

Answer

G = 79.99 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d = 37.47 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 20000/151.8 = 131.7 N·m

Diameter for strength

T = 131.7 N·m = 131714 N·mm

d_τ = ∛(16·131714/(π·60)) = 22.36 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·131714/(π·39000·1.745 × 10⁻⁵))^1/4 = 37.47 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.36, 37.47) = 37.47 mm

Answer

d = 37.47 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: E (x = 8 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 = 26.67 kN, R₂ = 38.33 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 = 53.33, M_C = 86.67, M_D = 120, M_E = 153.3 kN·m, M_F = 76.67, M_G = 0

Answer

Mark E: M_E = 153.3 kN·m

6.Shear force & bending moment

Answer: x_c = 3.75 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 = 10·7.5 = 75 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·5 = 10·7.5·3.75

R_B = 281.3/5 = 56.25 kN

ΣF_y = 0: R_A = 75 − 56.25 = 18.75 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·18.75/10 = 3.75 m

Answer

x_c = 3.75 m

7.Centroid & second moment of area

Answer: r = 63.94 mm

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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

b = 88 mm, h = 163 mm

A = BH − bh = 100·175 − 88·163 = 3156 mm²

I = (BH³ − bh³)/12 = 1.29 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(1.29 × 10⁷/3156) = 63.94 mm

Answer

r = 63.94 mm

8.Centroid & second moment of area

Answer: I_x = 256.9 × 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 = (36000·150 − 3200·90)/(36000 − 3200) = 155.9 mm

Full plate about ȳ

120·300³/12 + 36000·−5.854² = 271.2 × 10⁶ mm⁴

Hole about ȳ

80·40³/12 + 3200·−65.85² = 14.3 × 10⁶ mm⁴

Subtract

I_x = 256.9 × 10⁶ mm⁴

Answer

I_x = 256.9 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 47.7 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.

Required radius

C = (100 + 70)/2 = 85 MPa

R = σ₁ − C = 135 − 85 = 50 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 15²) = 47.7 MPa

Answer

τ_xy = 47.7 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = −109.8 MPa

Transformation equation

τ_θ = −95·0.9397 + −60·0.342 = −109.8 MPa

Answer

τ_θ = −109.8 MPa

11.Beam deflection

Answer: δ_P = 22.69 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 = 28000 N, a = 1000 mm, b = 7000 mm, L = 8000 mm

EI = 2.52 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 28000·1000²·7000²/(3·2.52 × 10¹²·8000) = 22.69 mm

Answer

δ_P = 22.69 mm

12.Beam deflection

Answer: I_req = 10.85 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 3000/360 = 8.333 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·6·3000⁴/(384·70000·8.333) = 1.085 × 10⁷ mm⁴

I_req = 10.85 × 10⁶ mm⁴

Answer

I_req = 10.85 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 52.64 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 40·20³/12 = 26667 mm⁴

Effective length

KL = 0.5·2000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·26667/1000² = 52638 N = 52.64 kN

Answer

P_cr = 52.64 kN

14.Column buckling

Answer: d = 55.77 mm

Required critical load

P_cr = n·P = 2.5·85000 = 212500 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·2100²/(π³·200000))^(1/4) = 55.77 mm

Answer

d = 55.77 mm

Check: the slenderness KL/r is 150.6, so the column is slender and Euler applies.

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