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 = 120 kNd = 10 mmL = 2.5 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 largest normal stress σ_max in the bar.

P = 45 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø16 mm, 0.5 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 80 mm, inner diameter 56 mm, length 2.5 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.

T = 1500 N·md_o = 80 mmd_i = 56 mmL = 2.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 30 kWN = 3000 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries two point loads. Find the maximum bending moment M_max in the beam.

L = 5 mP₁ = 20 kN at 3 mP₂ = 60 kN at 3.5 m

M_max = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 10 mP = 80 kN at 8 mw = 3 kN/m from 0 to 10 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A hollow rectangular (box) section has outer width B = 100 mm, outer height H = 350 mm and a uniform wall thickness t = 8 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 100 mmH = 350 mmt = 8 mm

r = mm

8.Centroid & second moment of areaHard

A 100 mm × 150 mm plate has a rectangular cut-out 60 mm wide and 40 mm high, centred on the vertical centre line with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmcut-out: 60 × 40 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 0 MPa, σᵧ = −40 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 0 MPaσ_y = −40 MPaτ_xy = 60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −80 MPa, σᵧ = −10 MPa and τₓᵧ = 60 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 = −80 MPaσ_y = −10 MPaτ_xy = 60 MPaθ = 35°

σ_θ = 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 = 4 kN/mL = 3 mE = 70 GPa (aluminium)I = 12 × 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 = 25 kNw = 12 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 1.5 mE = 200 GPa (steel)b × h = 40 × 100 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 = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 13059 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·10²/4 = 78.54 mm²

σ = P/A = 120·1000/78.54 = 1528 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1528/117000 = 0.01306 = 13059 µε

Answer

ε = 13059 µε

2.Axial stress & strain

Answer: σ_max = 223.8 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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 63.66 MPa

σ₂ = 223.8 MPa

Largest stress

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

σ_max = 223.8 MPa

Answer

σ_max = 223.8 MPa

3.Torsion of shafts

Answer: φ = 0.8789 °

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(80⁴ − 56⁴)/32 = 3.056 × 10⁶ mm⁴

Angle of twist

T = 1500 N·m = 1.5 × 10⁶ N·mm

L = 2.5 m = 2500 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 1.5 × 10⁶·2500/(80000·3.056 × 10⁶) = 0.01534 rad

φ = 0.01534·180/π = 0.8789°

Answer

φ = 0.8789°

4.Torsion of shafts

Answer: d = 20.09 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 30000/314.2 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·60)) = 20.09 mm

Answer

d = 20.09 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_max = 81 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.

ΣM_A = 0: R_B·5 = 20·3 + 60·3.5

R_B = 270/5 = 54 kN

ΣF_y = 0: R_A = 80 − 54 = 26 kN

Moment under each load

Keep the part left of each load.

under P₁: M = R_A·3 = 26·3 = 78 kN·m

under P₂: M = R_B·1.5 = 54·1.5 = 81 kN·m

Largest value

M_max = 81 kN·m (under P₂)

Answer

M_max = 81 kN·m

6.Shear force & bending moment

Answer: M_max = 152 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 80·8 + 3·10·5

R_B = 790/10 = 79 kN

ΣF_y = 0: R_A = 110 − 79 = 31 kN

Where is V = 0?

V(8⁻) = R_A − w·8 = 7 kN

V(8⁺) = 7 − 80 = −73 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·8 − w·8²/2 = 31·8 − 3·8²/2 = 152 kN·m

Answer

M_max = 152 kN·m

It occurs at x = 8 m from A.

7.Centroid & second moment of area

Answer: r = 117.9 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 = 84 mm, h = 334 mm

A = BH − bh = 100·350 − 84·334 = 6944 mm²

I = (BH³ − bh³)/12 = 9.647 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(9.647 × 10⁷/6944) = 117.9 mm

Answer

r = 117.9 mm

8.Centroid & second moment of area

Answer: I_x = 26.36 × 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 = (15000·75 − 2400·52.5)/(15000 − 2400) = 79.29 mm

Full plate about ȳ

100·150³/12 + 15000·−4.286² = 28.4 × 10⁶ mm⁴

Hole about ȳ

60·40³/12 + 2400·−26.79² = 2.042 × 10⁶ mm⁴

Subtract

I_x = 26.36 × 10⁶ mm⁴

Answer

I_x = 26.36 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 63.25 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 = √(((0 − −40)/2)² + 60²) = 63.25 MPa

Answer

τ_max = 63.25 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −0.5891 MPa

Transformation equation

σ_θ = −45 + −35·0.342 + 60·0.9397 = −0.5891 MPa

Answer

σ_θ = −0.5891 MPa

11.Beam deflection

Answer: θ_A = 0.005357 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 = 4 N/mm, L = 3000 mm

EI = 8.4 × 10¹¹ N·mm²

End slope

θ_A = wL³/(24EI) = 4·3000³/(24·8.4 × 10¹¹) = 0.005357 rad

Answer

θ_A = 0.005357 rad

That is 0.3069°.

12.Beam deflection

Answer: δ_max = 2.422 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 = 25000 N, w = 12 N/mm, L = 2500 mm, EI = 5.88 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 25000·2500³/(48·5.88 × 10¹²) = 1.384 mm

δ_w = 5wL⁴/(384EI) = 5·12·2500⁴/(384·5.88 × 10¹²) = 1.038 mm

Add them

δ_max = 1.384 + 1.038 = 2.422 mm

Answer

δ_max = 2.422 mm

13.Column buckling

Answer: P_cr = 467.9 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/1500² = 467892 N = 467.9 kN

Answer

P_cr = 467.9 kN

14.Column buckling

Answer: σ_cr = 8.812 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 0.7·3000 = 2100 mm

KL/r = 2100/7.5 = 280

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σ_cr = 8.812 MPa

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