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 bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 60 mm, length 500 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 500 mm, E = 110 GPaδ = 0.5 mm

P = kN

2.Axial stress & strainHard

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

P = 80 kNsegment 1: Ø40 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid brass shaft 2 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 1500 N·mL = 2 mG = 39 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A hollow shaft transmits 5 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 5 kWN = 300 rpmd_o = 75 mmd_i = 60 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the shear force V_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 = 6 kN/m from 0 to 4 mx = 0.5 m

V_x = kN

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x₀ from A to the section where the bending moment is largest.

L = 6 mw = 20 kN/m from 3 to 6 m

x₀ = m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 260 mm wide with two upright legs, 200 mm high overall. Base and legs are all 20 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 260 mmh = 200 mmt = 20 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 170 mm wide and 24 mm thick, an overall depth of 300 mm and a web 8 mm thick. It carries a bending moment of 120 kN·m about the axis x. Find the size of the maximum bending stress σ_max in the section.

B = 170 mmH = 300 mmt_f = 24 mmt_w = 8 mmM = 120 kN·m

σ_max = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 70 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 20° 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 = 70 MPaσ_y = −10 MPaτ_xy = −10 MPaθ = 20°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −10 MPa, σᵧ = 20 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −10 MPaσ_y = 20 MPaτ_xy = −60 MPaθ = 60°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 3 m from support A. Find the deflection δ_P under the load.

P = 22 kNa = 3 mL = 5 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_P = 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 = 4 kNw = 5 kN/mL = 2.5 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_max = 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 = 2.5 mE = 70 GPa (aluminium)d = 30 mm

P_cr = kN

14.Column bucklingHard

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

L = 2 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 107.2 kN

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

A₁ = 490.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(110000·2827) = 1.608 × 10⁻⁶ mm/N

Load

P = 0.5/4.663 × 10⁻⁶ = 107218 N = 107.2 kN

Answer

P = 107.2 kN

2.Axial stress & strain

Answer: δ = 5.464 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 = 63.66 MPa

σ₂ = 707.4 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 63.66·500/200000 = 0.1592 mm

δ₂ = σ₂L₂/E₂ = 707.4·1500/200000 = 5.305 mm

δ = δ₁ + δ₂ = 5.464 mm

Answer

δ = 5.464 mm

3.Torsion of shafts

Answer: d = 97.34 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 = 1500 N·m = 1.5 × 10⁶ N·mm

L = 2 m = 2000 mm

G = 39 GPa = 39000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 1.5 × 10⁶·2000/(39000·0.008727) = 8.815 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 97.34 mm

Answer

d = 97.34 mm

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

4.Torsion of shafts

Answer: τ_max = 3.254 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 5000/31.42 = 159.2 N·m

Polar second moment of area

J = π(75⁴ − 60⁴)/32 = 1.834 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 159155·37.5/1.834 × 10⁶ = 3.254 MPa

Answer

τ_max = 3.254 MPa

5.Shear force & bending moment

Answer: V_x = 9 kN

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·4 = 24 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 6·4·2

R_B = 48/4 = 12 kN

ΣF_y = 0: R_A = 24 − 12 = 12 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.

V = R_A − (all downward loads left of the cut)

V_x = 12 − 6·0.5 = 9 kN

Answer

V_x = 9 kN

6.Shear force & bending moment

Answer: x₀ = 3.75 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 = 20·3 = 60 kN, acting at x̄ = 4.5 m

ΣM_A = 0: R_B·6 = 20·3·4.5

R_B = 270/6 = 45 kN

ΣF_y = 0: R_A = 60 − 45 = 15 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 3) = 0

x₀ = 3 + 15/20 = 3.75 m

Answer

x₀ = 3.75 m

7.Centroid & second moment of area

Answer: I_x = 49.81 × 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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 260·20 = 5200 mm², y₁ = 10 mm

two legs: A₂ = 2·20·180 = 7200 mm², y₂ = 20 + 180/2 = 110 mm

Centroid

ȳ = (5200·10 + 7200·110)/12400 = 68.06 mm

Parallel-axis theorem

base: 260·20³/12 + 5200·(10 − 68.06)² = 1.771 × 10⁷ mm⁴

two legs: 2·20·180³/12 + 7200·(110 − 68.06)² = 3.21 × 10⁷ mm⁴

I_x = 1.771 × 10⁷ + 3.21 × 10⁷

I_x = 4.981 × 10⁷ mm⁴ = 49.81 × 10⁶ mm⁴

Answer

I_x = 49.81 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_max = 108.1 MPa

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 − 2·24 = 252 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (170·300³ − 162·252³)/12

I_x = 1.665 × 10⁸ mm⁴

Bending stress

y = 150 mm

σ = M·y/I = 1.2 × 10⁸·150/1.665 × 10⁸ = 108.1 MPa

Answer

σ_max = 108.1 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 5.786 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′ = 30 − 40·0.766 − (−10)·0.6428 = 5.786 MPa

Check

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

σ_x′ = 54.21 MPa, σ_x′ + σ_y′ = 60 = σ_x + σ_y

Answer

σ_y′ = 5.786 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 42.99 MPa

Transformation equation

τ_θ = −−15·0.866 + −60·−0.5 = 42.99 MPa

Answer

τ_θ = 42.99 MPa

11.Beam deflection

Answer: δ_P = 40 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.

P = 22000 N, a = 3000 mm, b = 2000 mm, L = 5000 mm

EI = 1.32 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 22000·3000²·2000²/(3·1.32 × 10¹²·5000) = 40 mm

Answer

δ_P = 40 mm

12.Beam deflection

Answer: δ_max = 1.602 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 = 4000 N, w = 5 N/mm, L = 2500 mm, EI = 2.4 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 4000·2500³/(48·2.4 × 10¹²) = 0.5425 mm

δ_w = 5wL⁴/(384EI) = 5·5·2500⁴/(384·2.4 × 10¹²) = 1.06 mm

Add them

δ_max = 0.5425 + 1.06 = 1.602 mm

Answer

δ_max = 1.602 mm

13.Column buckling

Answer: P_cr = 4.395 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·39761/2500² = 4395 N = 4.395 kN

Answer

P_cr = 4.395 kN

14.Column buckling

Answer: σ_cr = 49.35 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·2000 = 1000 mm

KL/r = 1000/5 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σ_cr = 49.35 MPa

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