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 elongation δ of the brass bar under the axial load.

P = 55 kNd = 30 mmL = 1 mE = 100 GPa

δ = mm

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 = 50 kNsegment 1: Ø30 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 60 mm and length 1 m; segment BC has diameter 30 mm and length 1 m. A torque of 750 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 750 N·md₁ = 60 mm, L₁ = 1 md₂ = 30 mm, L₂ = 1 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 500 rpmd_o = 75 mmd_i = 60 mm

τ_max = MPa

5.Shear force & bending momentMedium

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2.5 m from A. Find the bending moment M_x at the section x = 1.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 3 mP = 75 kN at 2.5 mw = 4 kN/m from 0 to 3 mx = 1.5 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 7.5 mw = 10 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m

x_c = m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 280 mm and inside diameter d = 252 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 280 mmd = 252 mm

r = mm

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 20 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 280 mmt_f = 30 mmt_w = 12 mmh_w = 120 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = 10 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 130 MPaσ_y = 10 MPaτ_xy = −60 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −30 MPa, σᵧ = 30 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σ_x = −30 MPaσ_y = 30 MPaτ_xy = −50 MPaθ = 20°

σ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 12 kN/mL = 6 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴

θ_A = rad

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 = 28 kNw = 20 kN/mL = 6 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 3 mE = 70 GPa (aluminium)b × h = 20 × 60 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.7781 mm

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 = 55·1000/706.9 = 77.81 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 77.81/100000 = 0.0007781 = 778.1 µε

Elongation

δ = ε·L = 0.0007781·1000 mm = 0.7781 mm

Answer

δ = 0.7781 mm

2.Axial stress & strain

Answer: σ_max = 248.7 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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 70.74 MPa

σ₂ = 248.7 MPa

Largest stress

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

σ_max = 248.7 MPa

Answer

σ_max = 248.7 MPa

3.Torsion of shafts

Answer: τ_max = 141.5 MPa

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.

Stress in each segment

T = 750 N·m = 750000 N·mm

τ_AB = 16·750000/(π·60³) = 17.68 MPa

τ_BC = 16·750000/(π·30³) = 141.5 MPa

Answer

τ_max = 141.5 MPa

4.Torsion of shafts

Answer: τ_max = 29.29 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 75000/52.36 = 1432 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 = 1.432 × 10⁶·37.5/1.834 × 10⁶ = 29.29 MPa

Answer

τ_max = 29.29 MPa

5.Shear force & bending moment

Answer: M_x = −79.5 kN·m

Keep the free-end part

P: −75·1 = −75 kN·m

w: −4·1.5²/2 = −4.5 kN·m

M_x = −75 − 4.5 = −79.5 kN·m

Answer

M_x = −79.5 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: x_c = 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 10·7.5 = 75 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·5 = 10·7.5·3.75

R_B = 281.3/5 = 56.25 kN

ΣF_y = 0: R_A = 75 − 56.25 = 18.75 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·18.75/10 = 3.75 m

Answer

x_c = 3.75 m

7.Centroid & second moment of area

Answer: r = 94.18 mm

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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 11699 mm²

I = π(D⁴ − d⁴)/64 = 1.038 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(1.038 × 10⁸/11699) = 94.18 mm

Answer

r = 94.18 mm

8.Centroid & second moment of area

Answer: σ_bottom = 267.5 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 = 280·30 = 8400 mm², y_f = 120 + 30/2 = 135 mm

web: A_w = 12·120 = 1440 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (8400·135 + 1440·60)/9840 = 124 mm

Parallel-axis theorem

flange: 280·30³/12 + 8400·(135 − 124)² = 1.642 × 10⁶ mm⁴

web: 12·120³/12 + 1440·(60 − 124)² = 7.631 × 10⁶ mm⁴

I_x = 1.642 × 10⁶ + 7.631 × 10⁶ = 9.273 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 124 mm

σ = M·y/I = 2 × 10⁷·124/9.273 × 10⁶ = 267.5 MPa

Answer

σ_bottom = 267.5 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −22.5 °

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.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−60, 130 − 10) = −45°

Angle on the element

θ_p = −22.5°

Answer

θ_p = −22.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = −55.12 MPa

Transformation equation

σ_θ = 0 + −30·0.766 + −50·0.6428 = −55.12 MPa

Answer

σ_θ = −55.12 MPa

11.Beam deflection

Answer: θ_A = 0.0045 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 12 N/mm, L = 6000 mm

EI = 2.4 × 10¹³ N·mm²

End slope

θ_A = wL³/(24EI) = 12·6000³/(24·2.4 × 10¹³) = 0.0045 rad

Answer

θ_A = 0.0045 rad

That is 0.2578°.

12.Beam deflection

Answer: δ_max = 27.59 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 = 28000 N, w = 20 N/mm, L = 6000 mm, EI = 1.68 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 28000·6000³/(48·1.68 × 10¹³) = 7.5 mm

δ_w = 5wL⁴/(384EI) = 5·20·6000⁴/(384·1.68 × 10¹³) = 20.09 mm

Add them

δ_max = 7.5 + 20.09 = 27.59 mm

Answer

δ_max = 27.59 mm

13.Column buckling

Answer: P_cr = 12.28 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 = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/1500² = 12282 N = 12.28 kN

Answer

P_cr = 12.28 kN

14.Column buckling

Answer: σ_cr = 1.079 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 2·5000 = 10000 mm

KL/r = 10000/12.5 = 800

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/800² = 1.079 MPa

Answer

σ_cr = 1.079 MPa

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