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

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 400 mm

δ = mm

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 = 85 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid brass shaft 1 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 1000 N·mL = 1 mG = 39 GPaφ_allow = 1°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 75 kW at 1450 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 75 kWN = 1450 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 5.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 6 kN/m from 0 to 6 mx = 5.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 60 kN at 5 mw = 2 kN/m from 0 to 8 mx = 5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 30 mmhole centre: y = 45 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 150 mm plate has a rectangular cut-out 40 mm wide and 50 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 = 200 mmh = 150 mmcut-out: 40 × 50 mmhole centre: y = 90 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 140 MPa, σᵧ = 80 MPa and τₓᵧ = −50 MPa (tension positive). The element is rotated 30° 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 = 140 MPaσ_y = 80 MPaτ_xy = −50 MPaθ = 30°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 90 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 90 MPaσ_y = −10 MPaτ_xy = 30 MPaθ = 50°

τ_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 = 5 mE = 200 GPa (steel)I = 84 × 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 = 6 kN/mL = 4 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 pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 200 GPa (steel)b × h = 20 × 40 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.05628 mm

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·20²/4 = 314.2 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·5890 = 7.108 × 10⁸ N

Shortening

δ = 100000·400/7.108 × 10⁸ = 0.05628 mm

Answer

δ = 0.05628 mm

2.Axial stress & strain

Answer: σ_max = 422.8 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·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 88.35 MPa

σ₂ = 422.8 MPa

Largest stress

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

σ_max = 422.8 MPa

Answer

σ_max = 422.8 MPa

3.Torsion of shafts

Answer: d = 62.2 mm

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.

Required polar moment

T = 1000 N·m = 1 × 10⁶ N·mm

L = 1 m = 1000 mm

G = 39 GPa = 39000 MPa

φ = 1·π/180 = 0.01745 rad

J = TL/(Gφ) = 1 × 10⁶·1000/(39000·0.01745) = 1.469 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 62.2 mm

Answer

d = 62.2 mm

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

4.Torsion of shafts

Answer: d = 36.92 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 75000/151.8 = 493.9 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·493929/(π·50)) = 36.92 mm

Answer

d = 36.92 mm

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

5.Shear force & bending moment

Answer: M_x = 8.25 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 = 6·6 = 36 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 6·6·3

R_B = 108/6 = 18 kN

ΣF_y = 0: R_A = 36 − 18 = 18 kN

Cut at x = 5.5 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 = 18·5.5 − 6·5.5·2.75 = 8.25 kN·m

Answer

M_x = 8.25 kN·m

6.Shear force & bending moment

Answer: M_x = 127.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·8 = 16 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 60·5 + 2·8·4

R_B = 364/8 = 45.5 kN

ΣF_y = 0: R_A = 76 − 45.5 = 30.5 kN

Cut at x = 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 = 30.5·5 − 2·5·2.5 = 127.5 kN·m

Answer

M_x = 127.5 kN·m

7.Centroid & second moment of area

Answer: I_x = 27.42 × 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 − 706.9·45)/(15000 − 706.9) = 76.48 mm

Full plate about ȳ

100·150³/12 + 15000·−1.484² = 28.16 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−31.48² = 0.7404 × 10⁶ mm⁴

Subtract

I_x = 27.42 × 10⁶ mm⁴

Answer

I_x = 27.42 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 55.35 × 10⁶ mm⁴

Centroid

ȳ = (30000·75 − 2000·90)/(30000 − 2000) = 73.93 mm

Full plate about ȳ

200·150³/12 + 30000·1.071² = 56.28 × 10⁶ mm⁴

Hole about ȳ

40·50³/12 + 2000·16.07² = 0.9332 × 10⁶ mm⁴

Subtract

I_x = 55.35 × 10⁶ mm⁴

Answer

I_x = 55.35 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 138.3 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.

Transformation equation for the y′ face

σ_y′ = 110 − 30·0.5 − (−50)·0.866 = 138.3 MPa

Check

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

σ_x′ = 81.7 MPa, σ_x′ + σ_y′ = 220 = σ_x + σ_y

Answer

σ_y′ = 138.3 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = −54.45 MPa

Transformation equation

τ_θ = −50·0.9848 + 30·−0.1736 = −54.45 MPa

Answer

τ_θ = −54.45 MPa

11.Beam deflection

Answer: δ_max = 2.422 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 = 5000 mm

EI = 1.68 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·5·5000⁴/(384·1.68 × 10¹³) = 2.422 mm

Answer

δ_max = 2.422 mm

12.Beam deflection

Answer: I_req = 9 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·6·4000⁴/(384·200000·11.11) = 9 × 10⁶ mm⁴

I_req = 9 × 10⁶ mm⁴

Answer

I_req = 9 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 107.4 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 = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·26667/700² = 107424 N = 107.4 kN

Answer

P_cr = 107.4 kN

14.Column buckling

Answer: σ_cr = 4.819 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 1·4000 = 4000 mm

KL/r = 4000/6.25 = 640

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/640² = 4.819 MPa

Answer

σ_cr = 4.819 MPa

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