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 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 70 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø50 mm, E = 120 GPa, σ_allow = 50 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid steel 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 = 1000 N·md = 40 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 30 kWN = 1000 rpmτ_allow = 75 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 12 kN/m from 0 to 4 mx = 0.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 25 kN at 3 mw = 3 kN/m from 0 to 10 mx = 4.5 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 90 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 = 90 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 80 mm wide and 50 mm high, centred on the vertical centre line with its centre 156 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmcut-out: 80 × 50 mmhole centre: y = 156 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = 60 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 130 MPaσ_y = 60 MPaτ_xy = −10 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = 60 MPa and τₓᵧ = 50 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 30° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 10 MPaσ_y = 60 MPaτ_xy = 50 MPaθ = 30°

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 = 15 kN/mL = 2 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the deflection δ_mid at midspan.

P = 21 kNb = 1.5 mL = 4 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_mid = mm

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 = 3 mE = 200 GPa (steel)b × h = 20 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 145 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 145 kNn = 3L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.06205 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·40²/4 = 1257 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·7461 = 7.736 × 10⁸ N

Shortening

δ = 80000·600/7.736 × 10⁸ = 0.06205 mm

Answer

δ = 0.06205 mm

2.Axial stress & strain

Answer: P_max = 399.2 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·5890 = 9.582 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0004167) = 0.0004167 (copper)

Largest load

P_max = 0.0004167·9.582 × 10⁸ = 399244 N = 399.2 kN

Answer

P_max = 399.2 kN

3.Torsion of shafts

Answer: φ = 5.699 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·2000/(80000·251327) = 0.09947 rad

φ = 0.09947·180/π = 5.699°

Answer

φ = 5.699°

4.Torsion of shafts

Answer: d = 26.89 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 30000/104.7 = 286.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·286479/(π·75)) = 26.89 mm

Answer

d = 26.89 mm

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

5.Shear force & bending moment

Answer: M_x = 10.5 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 = 12·4 = 48 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 12·4·2

R_B = 96/4 = 24 kN

ΣF_y = 0: R_A = 48 − 24 = 24 kN

Cut at x = 0.5 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 = 24·0.5 − 12·0.5·0.25 = 10.5 kN·m

Answer

M_x = 10.5 kN·m

6.Shear force & bending moment

Answer: M_x = 78.38 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 = 25·3 + 3·10·5

R_B = 225/10 = 22.5 kN

ΣF_y = 0: R_A = 55 − 22.5 = 32.5 kN

Cut at x = 4.5 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 = 32.5·4.5 − 25·(4.5 − 3) − 3·4.5·2.25 = 78.38 kN·m

Answer

M_x = 78.38 kN·m

7.Centroid & second moment of area

Answer: I_x = 27.92 × 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 − 706.9·90)/(15000 − 706.9) = 74.26 mm

Full plate about ȳ

100·150³/12 + 15000·0.7418² = 28.13 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·15.74² = 0.2149 × 10⁶ mm⁴

Subtract

I_x = 27.92 × 10⁶ mm⁴

Answer

I_x = 27.92 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 131.4 × 10⁶ mm⁴

Centroid

ȳ = (28800·120 − 4000·156)/(28800 − 4000) = 114.2 mm

Full plate about ȳ

120·240³/12 + 28800·5.806² = 139.2 × 10⁶ mm⁴

Hole about ȳ

80·50³/12 + 4000·41.81² = 7.824 × 10⁶ mm⁴

Subtract

I_x = 131.4 × 10⁶ mm⁴

Answer

I_x = 131.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 36.4 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 = √(((130 − 60)/2)² + −10²) = 36.4 MPa

Answer

τ_max = 36.4 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 46.65 MPa

Transformation equation

τ_θ = −−25·0.866 + 50·0.5 = 46.65 MPa

Answer

τ_θ = 46.65 MPa

11.Beam deflection

Answer: δ_max = 1.776 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 = 15 N/mm, L = 2000 mm

EI = 1.76 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·15·2000⁴/(384·1.76 × 10¹²) = 1.776 mm

Answer

δ_max = 1.776 mm

12.Beam deflection

Answer: δ_mid = 19.39 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 = 21000 N, b = 1500 mm, L = 4000 mm

EI = 1.32 × 10¹² N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 21000·1500·(3·4000² − 4·1500²)/(48·1.32 × 10¹²) = 19.39 mm

Answer

δ_mid = 19.39 mm

13.Column buckling

Answer: P_cr = 17.9 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·40000/2100² = 17904 N = 17.9 kN

Answer

P_cr = 17.9 kN

14.Column buckling

Answer: d = 65.1 mm

Required critical load

P_cr = n·P = 3·145000 = 435000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·435000·2000²/(π³·200000))^(1/4) = 65.1 mm

Answer

d = 65.1 mm

Check: the slenderness KL/r is 122.9, so the column is slender and Euler applies.

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