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 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 800 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 800 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø80/Ø50 mm, E = 70 GPa, σ_allow = 100 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid aluminium shaft 0.5 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 750 N·mL = 0.5 mG = 26 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 1800 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1800 rpmτ_allow = 80 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 40 kN·m at x = 4 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM₀ = 40 kN·m clockwise, at 4 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 60 kN at 7 mw = 10 kN/m from 0 to 8 mx = 3 m

M_x = kN·m

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 120 mmt_f = 15 mmt_w = 20 mmh_w = 100 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 240 mm plate has a rectangular cut-out 60 mm wide and 20 mm high, centred on the vertical centre line with its centre 84 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 240 mmcut-out: 60 × 20 mmhole centre: y = 84 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 90 MPa, σᵧ = 90 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 90 MPaσ_y = 90 MPaτ_xy = −20 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 100 MPa, σᵧ = 30 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σ_x = 100 MPaσ_y = 30 MPaτ_xy = −10 MPaθ = 40°

σ_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 = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

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 = 22 kNw = 4 kN/mL = 3 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ_max = mm

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 = 3 mE = 200 GPa (steel)b × h = 50 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 10 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 10 kNn = 2.5L = 2.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.2315 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·30²/4 = 706.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Shortening

δ = 300000·800/1.037 × 10⁹ = 0.2315 mm

Answer

δ = 0.2315 mm

2.Axial stress & strain

Answer: P_max = 390.7 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·3063 = 3.126 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001429) = 0.00125 (steel)

Largest load

P_max = 0.00125·3.126 × 10⁸ = 390736 N = 390.7 kN

Answer

P_max = 390.7 kN

3.Torsion of shafts

Answer: d = 64.05 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 = 750 N·m = 750000 N·mm

L = 0.5 m = 500 mm

G = 26 GPa = 26000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 750000·500/(26000·0.008727) = 1.653 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 64.05 mm

Answer

d = 64.05 mm

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

4.Torsion of shafts

Answer: d = 17.17 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 15000/188.5 = 79.58 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·79577/(π·80)) = 17.17 mm

Answer

d = 17.17 mm

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

5.Shear force & bending moment

Answer: M_x = −16 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 40/10 = 4 kN (upward)

R_A = −R_B = −4 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −4·4 = −16 kN·m

Answer

M_x = −16 kN·m

6.Shear force & bending moment

Answer: M_x = 97.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·8 = 80 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 60·7 + 10·8·4

R_B = 740/8 = 92.5 kN

ΣF_y = 0: R_A = 140 − 92.5 = 47.5 kN

Cut at x = 3 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 = 47.5·3 − 10·3·1.5 = 97.5 kN·m

Answer

M_x = 97.5 kN·m

7.Centroid & second moment of area

Answer: I_x = 4.833 × 10⁶ mm⁴

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 120·15 = 1800 mm², y_f = 100 + 15/2 = 107.5 mm

web: A_w = 20·100 = 2000 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1800·107.5 + 2000·50)/3800 = 77.24 mm

Parallel-axis theorem

flange: 120·15³/12 + 1800·(107.5 − 77.24)² = 1.682 × 10⁶ mm⁴

web: 20·100³/12 + 2000·(50 − 77.24)² = 3.15 × 10⁶ mm⁴

I_x = 1.682 × 10⁶ + 3.15 × 10⁶ = 4.833 × 10⁶ mm⁴ = 4.833 × 10⁶ mm⁴

Answer

I_x = 4.833 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 171.2 × 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 = (36000·120 − 1200·84)/(36000 − 1200) = 121.2 mm

Full plate about ȳ

150·240³/12 + 36000·−1.241² = 172.9 × 10⁶ mm⁴

Hole about ȳ

60·20³/12 + 1200·−37.24² = 1.704 × 10⁶ mm⁴

Subtract

I_x = 171.2 × 10⁶ mm⁴

Answer

I_x = 171.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 20 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 = √(((90 − 90)/2)² + −20²) = 20 MPa

Answer

τ_max = 20 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 61.23 MPa

Transformation equation

σ_θ = 65 + 35·0.1736 + −10·0.9848 = 61.23 MPa

Answer

σ_θ = 61.23 MPa

11.Beam deflection

Answer: δ_max = 1.838 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 = 1.7 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·15·2000⁴/(384·1.7 × 10¹²) = 1.838 mm

Answer

δ_max = 1.838 mm

12.Beam deflection

Answer: δ_max = 17.96 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 = 22000 N, w = 4 N/mm, L = 3000 mm, EI = 9.24 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 22000·3000³/(48·9.24 × 10¹¹) = 13.39 mm

δ_w = 5wL⁴/(384EI) = 5·4·3000⁴/(384·9.24 × 10¹¹) = 4.566 mm

Add them

δ_max = 13.39 + 4.566 = 17.96 mm

Answer

δ_max = 17.96 mm

13.Column buckling

Answer: P_cr = 548.3 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 = 60·50³/12 = 625000 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·625000/1500² = 548311 N = 548.3 kN

Answer

P_cr = 548.3 kN

14.Column buckling

Answer: d = 50.4 mm

Required critical load

P_cr = n·P = 2.5·10000 = 25000 N

KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·25000·5000²/(π³·200000))^(1/4) = 50.4 mm

Answer

d = 50.4 mm

Check: the slenderness KL/r is 396.9, so the column is slender and Euler applies.

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