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 aluminium bar, in microstrain (µε = 10⁻⁶).

P = 120 kNd = 30 mmL = 0.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 15 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel shaft 2 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 750 N·mL = 2 mG = 80 GPaφ_allow = 2°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 1800 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 20 kWN = 1800 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 = 12 kN/m from 0 to 6 mx = 5.5 m

M_x = kN·m

6.Shear force & bending momentHard

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

45 kN @ 6 m40 kN @ 1 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 140 mm wide and 12 mm thick, an overall depth of 260 mm and a web 8 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 140 mmH = 260 mmt_f = 12 mmt_w = 8 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 15 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 180 mmt_f = 25 mmt_w = 14 mmh_w = 100 mmM = 15 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 20 MPa and σᵧ = 50 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 110 MPa?

σ_x = 20 MPaσ_y = 50 MPaσ₁ = 110 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 40 MPa, σᵧ = 100 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 = 40 MPaσ_y = 100 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 = 5 kN/mL = 3 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 150 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 1.5 kN/mL = 5 mE = 11 GPa (timber)b = 150 mm

h_req = mm

13.Column bucklingMedium

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

L = 2 mE = 70 GPa (aluminium)b × h = 20 × 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 35 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 35 kNn = 2.5L = 1 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 2425 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·30²/4 = 706.9 mm²

σ = P/A = 120·1000/706.9 = 169.8 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 169.8/70000 = 0.002425 = 2425 µε

Answer

ε = 2425 µε

2.Axial stress & strain

Answer: δ = 2.927 mm

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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 11.94 MPa

σ₂ = 132.6 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 11.94·500/70000 = 0.08526 mm

δ₂ = σ₂L₂/E₂ = 132.6·1500/70000 = 2.842 mm

δ = δ₁ + δ₂ = 2.927 mm

Answer

δ = 2.927 mm

3.Torsion of shafts

Answer: d = 48.36 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 = 2 m = 2000 mm

G = 80 GPa = 80000 MPa

φ = 2·π/180 = 0.03491 rad

J = TL/(Gφ) = 750000·2000/(80000·0.03491) = 537148 mm⁴

Diameter

d = (32J/π)^1/4 = 48.36 mm

Answer

d = 48.36 mm

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

4.Torsion of shafts

Answer: d = 22.11 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 20000/188.5 = 106.1 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·106103/(π·50)) = 22.11 mm

Answer

d = 22.11 mm

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

5.Shear force & bending moment

Answer: M_x = 16.5 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 = 12·6 = 72 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 12·6·3

R_B = 216/6 = 36 kN

ΣF_y = 0: R_A = 72 − 36 = 36 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 = 36·5.5 − 12·5.5·2.75 = 16.5 kN·m

Answer

M_x = 16.5 kN·m

6.Shear force & bending moment

Answer: E (x = 4 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 7.5 kN, R₂ = 77.5 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 7.5, M_C = −25, M_D = −57.5, M_E = −90 kN·m, M_F = −45, M_G = 0

Answer

Mark E: M_E = −90 kN·m

7.Centroid & second moment of area

Answer: I_x = 60.47 × 10⁶ mm⁴

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.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 260 − 2·12 = 236 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (140·260³ − 132·236³)/12

I_x = 6.047 × 10⁷ mm⁴ = 60.47 × 10⁶ mm⁴

Answer

I_x = 60.47 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 262.9 MPa

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 = 180·25 = 4500 mm², y_f = 100 + 25/2 = 112.5 mm

web: A_w = 14·100 = 1400 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4500·112.5 + 1400·50)/5900 = 97.67 mm

Parallel-axis theorem

flange: 180·25³/12 + 4500·(112.5 − 97.67)² = 1.224 × 10⁶ mm⁴

web: 14·100³/12 + 1400·(50 − 97.67)² = 4.348 × 10⁶ mm⁴

I_x = 1.224 × 10⁶ + 4.348 × 10⁶ = 5.572 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 97.67 mm

σ = M·y/I = 1.5 × 10⁷·97.67/5.572 × 10⁶ = 262.9 MPa

Answer

σ_bottom = 262.9 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 73.48 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.

Required radius

C = (20 + 50)/2 = 35 MPa

R = σ₁ − C = 110 − 35 = 75 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(75² − 15²) = 73.48 MPa

Answer

τ_xy = 73.48 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 54.94 MPa

Transformation equation

σ_θ = 70 + −30·0.1736 + −10·0.9848 = 54.94 MPa

Answer

σ_θ = 54.94 MPa

11.Beam deflection

Answer: δ_max = 2.093 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 = 3000 mm

EI = 2.52 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·5·3000⁴/(384·2.52 × 10¹²) = 2.093 mm

Answer

δ_max = 2.093 mm

12.Beam deflection

Answer: h_req = 185.6 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.

w = 1.5 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1.5·5000⁴/(384·11000·13.89) = 7.99 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·7.99 × 10⁷/150)^1/3 = 185.6 mm

Answer

h_req = 185.6 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 5.757 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·33333/2000² = 5757 N = 5.757 kN

Answer

P_cr = 5.757 kN

14.Column buckling

Answer: d = 30.83 mm

Required critical load

P_cr = n·P = 2.5·35000 = 87500 N

KL = 1·1000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·87500·1000²/(π³·200000))^(1/4) = 30.83 mm

Answer

d = 30.83 mm

Check: the slenderness KL/r is 129.8, so the column is slender and Euler applies.

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