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

P = 30 kNd = 16 mmL = 2.5 mE = 200 GPa

ε = µε

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 = 40 kNsegment 1: Ø35 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø20 mm, 1.5 m, brass (E = 100 GPa)

σ_max = MPa

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 = 200 N·md = 30 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1000 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

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.

10 kN @ 2 m12 kN/m, 0–12 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 8 mP = 20 kN at 6 mw = 6 kN/m from 0 to 8 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 140 mm wide and 22 mm thick, an overall depth of 420 mm and a web 14 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 140 mmH = 420 mmt_f = 22 mmt_w = 14 mm

S_x = × 10³ mm³

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 200 mm wide and 18 mm thick, an overall depth of 300 mm and a web 6 mm thick. It carries a bending moment of 120 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 200 mmH = 300 mmt_f = 18 mmt_w = 6 mmM = 120 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 10 MPa, σᵧ = 10 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 10 MPaσ_y = 10 MPaτ_xy = −50 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = 100 MPa and τₓᵧ = 50 MPa (σ_z = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 160 MPaσ_y = 100 MPaτ_xy = 50 MPaσ_Y = 250 MPa

n =

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 = 4 kN/mL = 6 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1 m from support B. Find the deflection δ_mid at midspan.

P = 15 kNb = 1 mL = 8 mE = 70 GPa (aluminium)I = 12 × 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 = 2 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 150 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 150 kNn = 2L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 746 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 = 30·1000/201.1 = 149.2 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 149.2/200000 = 0.000746 = 746 µε

Answer

ε = 746 µε

2.Axial stress & strain

Answer: σ_max = 127.3 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·20²/4 = 314.2 mm²

σ₁ = 41.58 MPa

σ₂ = 127.3 MPa

Largest stress

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

σ_max = 127.3 MPa

Answer

σ_max = 127.3 MPa

3.Torsion of shafts

Answer: φ = 2.162 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·1200/(80000·79522) = 0.03773 rad

φ = 0.03773·180/π = 2.162°

Answer

φ = 2.162°

4.Torsion of shafts

Answer: d = 34.35 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 50000/104.7 = 477.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·477465/(π·60)) = 34.35 mm

Answer

d = 34.35 mm

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

5.Shear force & bending moment

Answer: D (x = 6 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 = 80.33 kN, R₂ = 73.67 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 = 136.7, M_C = 205.3, M_D = 226 kN·m, M_E = 198.7, M_F = 123.3, M_G = 0

Answer

Mark D: M_D = 226 kN·m

6.Shear force & bending moment

Answer: M_max = 70.08 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 20·6 + 6·8·4

R_B = 312/8 = 39 kN

ΣF_y = 0: R_A = 68 − 39 = 29 kN

Where is V = 0?

R_A − w·x₀ = 0 → x₀ = 29/6 = 4.833 m

Maximum moment

M_max = R_A·x₀ − w·x₀²/2 = 29·4.833 − 6·4.833²/2 = 70.08 kN·m

Answer

M_max = 70.08 kN·m

It occurs at x = 4.833 m from A.

7.Centroid & second moment of area

Answer: S_x = 1458 × 10³ mm³

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 = 420 − 2·22 = 376 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (140·420³ − 126·376³)/12

I_x = 3.062 × 10⁸ mm⁴

Section modulus

c = H/2 = 210 mm

S_x = I/c = 3.062 × 10⁸/210

S_x = 1.458 × 10⁶ mm³ = 1458 × 10³ mm³

Answer

S_x = 1458 × 10³ mm³

8.Centroid & second moment of area

Answer: σ_j = 103.8 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.

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 = 300 − 2·18 = 264 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (200·300³ − 194·264³)/12

I_x = 1.525 × 10⁸ mm⁴

Bending stress

y = 150 − 18 = 132 mm

σ = M·y/I = 1.2 × 10⁸·132/1.525 × 10⁸ = 103.8 MPa

Answer

σ_j = 103.8 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 50 MPa

Radius of Mohr’s circle

τ_max = R = √(((10 − 10)/2)² + −50²) = 50 MPa

Answer

τ_max = 50 MPa

10.Mohr’s circle & stress transformation

Answer: n = 1.328

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 = 130 MPa, R = 58.31 MPa

σ₁ = 188.3 MPa, σ₂ = 71.69 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 188.3 MPa

n = 250/188.3 = 1.328

Answer

n = 1.328

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_max = 28.13 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 = 4 N/mm, L = 6000 mm

EI = 2.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·6000⁴/(384·2.4 × 10¹²) = 28.13 mm

Answer

δ_max = 28.13 mm

12.Beam deflection

Answer: δ_mid = 69.94 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 = 15000 N, b = 1000 mm, L = 8000 mm

EI = 8.4 × 10¹¹ N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 15000·1000·(3·8000² − 4·1000²)/(48·8.4 × 10¹¹) = 69.94 mm

Answer

δ_mid = 69.94 mm

13.Column buckling

Answer: P_cr = 309 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 round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/1400² = 308975 N = 309 kN

Answer

P_cr = 309 kN

14.Column buckling

Answer: d = 102 mm

Required critical load

P_cr = n·P = 2·150000 = 300000 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·3500²/(π³·70000))^(1/4) = 102 mm

Answer

d = 102 mm

Check: the slenderness KL/r is 137.2, so the column is slender and Euler applies.

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