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 = 70 kNd = 16 mmL = 3 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø80/Ø60 mm, E = 110 GPa, σ_allow = 50 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid aluminium 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 = 100 N·md = 20 mmL = 1 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1800 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1800 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

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

L = 5 mw = 8 kN/m from 0 to 5 mx = 4 m

V_x = kN

6.Shear force & bending momentHard

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 mP = 55 kN at 2.5 mw = 12 kN/m from 0 to 5 mx = 2.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A hollow rectangular (box) section has outer width B = 240 mm, outer height H = 400 mm and a uniform wall thickness t = 14 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 240 mmH = 400 mmt = 14 mm

r = mm

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 180 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 × 20 mmhole centre: y = 180 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −80 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −80 MPaσ_y = 50 MPaτ_xy = 30 MPaθ = 25°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 40 MPa, σᵧ = 30 MPa and τₓᵧ = 20 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 40 MPaσ_y = 30 MPaτ_xy = 20 MPa

σ_vM = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_B at the free end B.

P = 27 kNa = 1 mL = 2 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_B = 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 = 12 kN/mL = 4 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

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 = 2 mE = 200 GPa (steel)b × h = 40 × 100 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 = 4 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 2976 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 = 70·1000/201.1 = 348.2 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 348.2/117000 = 0.002976 = 2976 µε

Answer

ε = 2976 µε

2.Axial stress & strain

Answer: P_max = 174.2 kN

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·2199 = 3.833 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0004545) = 0.0004545 (bronze)

Largest load

P_max = 0.0004545·3.833 × 10⁸ = 174216 N = 174.2 kN

Answer

P_max = 174.2 kN

3.Torsion of shafts

Answer: φ = 14.03 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1000/(26000·15708) = 0.2449 rad

φ = 0.2449·180/π = 14.03°

Answer

φ = 14.03°

4.Torsion of shafts

Answer: d = 37.81 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 100000/188.5 = 530.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·530516/(π·50)) = 37.81 mm

Answer

d = 37.81 mm

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

5.Shear force & bending moment

Answer: V_x = −12 kN

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 = 8·5 = 40 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 8·5·2.5

R_B = 100/5 = 20 kN

ΣF_y = 0: R_A = 40 − 20 = 20 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.

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

V_x = 20 − 8·4 = −12 kN

Answer

V_x = −12 kN

6.Shear force & bending moment

Answer: M_x = 106.3 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·5 = 60 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 55·2.5 + 12·5·2.5

R_B = 287.5/5 = 57.5 kN

ΣF_y = 0: R_A = 115 − 57.5 = 57.5 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 = 57.5·2.5 − 12·2.5·1.25 = 106.3 kN·m

Answer

M_x = 106.3 kN·m

7.Centroid & second moment of area

Answer: r = 147 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 = 212 mm, h = 372 mm

A = BH − bh = 240·400 − 212·372 = 17136 mm²

I = (BH³ − bh³)/12 = 3.705 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(3.705 × 10⁸/17136) = 147 mm

Answer

r = 147 mm

8.Centroid & second moment of area

Answer: I_x = 268.4 × 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 − 1600·180)/(36000 − 1600) = 148.6 mm

Full plate about ȳ

120·300³/12 + 36000·1.395² = 270.1 × 10⁶ mm⁴

Hole about ȳ

80·20³/12 + 1600·31.4² = 1.63 × 10⁶ mm⁴

Subtract

I_x = 268.4 × 10⁶ mm⁴

Answer

I_x = 268.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 3.8 MPa

Transformation equation for the y′ face

σ_y′ = −15 − (−65)·0.6428 − 30·0.766 = 3.8 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −33.8 MPa, σ_x′ + σ_y′ = −30 = σ_x + σ_y

Answer

σ_y′ = 3.8 MPa

10.Mohr’s circle & stress transformation

Answer: σ_vM = 50 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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 35 MPa, R = 20.62 MPa

σ₁ = 55.62 MPa, σ₂ = 14.38 MPa, σ₃ = 0

Von Mises

σ_vM = √(55.62² − 55.62·14.38 + 14.38²) = 50 MPa

Answer

σ_vM = 50 MPa

11.Beam deflection

Answer: δ_B = 5.639 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 = 27000 N, a = 1000 mm, L = 2000 mm

EI = 3.99 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 2.256 mm

θ_P = Pa²/(2EI) = 0.003383 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 5.639 mm

Answer

δ_B = 5.639 mm

12.Beam deflection

Answer: I_req = 18 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·12·4000⁴/(384·200000·11.11) = 1.8 × 10⁷ mm⁴

I_req = 18 × 10⁶ mm⁴

Answer

I_req = 18 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 263.2 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/2000² = 263189 N = 263.2 kN

Answer

P_cr = 263.2 kN

14.Column buckling

Answer: σ_cr = 17.27 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 = 1·4000 = 4000 mm

KL/r = 4000/20 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa

Answer

σ_cr = 17.27 MPa

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