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 aluminium bar under the axial load.

P = 85 kNd = 20 mmL = 3 mE = 70 GPa

δ = mm

2.Axial stress & strainHard

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

P = 45 kNsegment 1: Ø40 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid aluminium 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 = 100 N·md = 20 mmL = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft of diameter 40 mm is fixed at A. A torque of 1500 N·m acts at B, 1.2 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT_B = 1500 N·mT_C = 200 N·mAB = 1.2 mBC = 1.2 m

τ_max = MPa

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 10 kN/m from 0 to 5 mx = 4 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.

20 kN @ 0 m55 kN @ 6 m10 kN @ 2 m8 kN/m, 1–5 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 150 mm × 200 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 200 mmd = 30 mmhole centre: y = 60 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 240 mm plate has a rectangular cut-out 80 mm wide and 40 mm high, centred on the vertical centre line with its centre 96 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: 80 × 40 mmhole centre: y = 96 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 35° 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 = −30 MPaσ_y = 100 MPaτ_xy = −10 MPaθ = 35°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −10 MPa, σᵧ = 90 MPa and τₓᵧ = 40 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 = −10 MPaσ_y = 90 MPaτ_xy = 40 MPaθ = 20°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 4 kN/mL = 2 mE = 11 GPa (timber)I = 12 × 10⁶ mm⁴

δ = 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 = 12 kNw = 5 kN/mL = 1.5 mE = 11 GPa (timber)I = 8.5 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 5 mE = 70 GPa (aluminium)d = 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 95 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 95 kNn = 2.5L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 11.6 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·20²/4 = 314.2 mm²

σ = P/A = 85·1000/314.2 = 270.6 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 270.6/70000 = 0.003865 = 3865 µε

Elongation

δ = ε·L = 0.003865·3000 mm = 11.6 mm

Answer

δ = 11.6 mm

2.Axial stress & strain

Answer: δ = 3.609 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 = 35.81 MPa

σ₂ = 397.9 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 35.81·1500/70000 = 0.7674 mm

δ₂ = σ₂L₂/E₂ = 397.9·500/70000 = 2.842 mm

δ = δ₁ + δ₂ = 3.609 mm

Answer

δ = 3.609 mm

3.Torsion of shafts

Answer: φ = 11.22 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·800/(26000·15708) = 0.1959 rad

φ = 0.1959·180/π = 11.22°

Answer

φ = 11.22°

4.Torsion of shafts

Answer: τ_max = 135.3 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 200 N·m

T_AB = T_C + T_B = 200 + 1500 = 1700 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 1700 N·m

τ_max = 16·1.7 × 10⁶/(π·40³) = 135.3 MPa

Answer

τ_max = 135.3 MPa

5.Shear force & bending moment

Answer: M_x = 20 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 = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 10·5·2.5

R_B = 125/5 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 4 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 = 25·4 − 10·4·2 = 20 kN·m

Answer

M_x = 20 kN·m

6.Shear force & bending moment

Answer: F (x = 5 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 = 34.75 kN, R₂ = 82.25 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 = −20, M_C = −9.25, M_D = −16.5, M_E = −31.75, M_F = −55 kN·m, M_G = 0

Answer

Mark F: M_F = −55 kN·m

7.Centroid & second moment of area

Answer: I_x = 98.8 × 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 = (30000·100 − 706.9·60)/(30000 − 706.9) = 101 mm

Full plate about ȳ

150·200³/12 + 30000·−0.9652² = 100 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−40.97² = 1.226 × 10⁶ mm⁴

Subtract

I_x = 98.8 × 10⁶ mm⁴

Answer

I_x = 98.8 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 170.4 × 10⁶ mm⁴

Centroid

ȳ = (36000·120 − 3200·96)/(36000 − 3200) = 122.3 mm

Full plate about ȳ

150·240³/12 + 36000·−2.341² = 173 × 10⁶ mm⁴

Hole about ȳ

80·40³/12 + 3200·−26.34² = 2.647 × 10⁶ mm⁴

Subtract

I_x = 170.4 × 10⁶ mm⁴

Answer

I_x = 170.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 66.63 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′ = 35 − (−65)·0.342 − (−10)·0.9397 = 66.63 MPa

Check

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

σ_x′ = 3.372 MPa, σ_x′ + σ_y′ = 70 = σ_x + σ_y

Answer

σ_y′ = 66.63 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 27.41 MPa

Transformation equation

σ_θ = 40 + −50·0.766 + 40·0.6428 = 27.41 MPa

Answer

σ_θ = 27.41 MPa

11.Beam deflection

Answer: δ = 60.61 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 = 2000 mm

EI = 1.32 × 10¹¹ N·mm²

Deflection

δ = wL⁴/(8EI) = 4·2000⁴/(8·1.32 × 10¹¹) = 60.61 mm

Answer

δ = 60.61 mm

12.Beam deflection

Answer: δ_max = 12.55 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 = 12000 N, w = 5 N/mm, L = 1500 mm, EI = 9.35 × 10¹⁰ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 12000·1500³/(48·9.35 × 10¹⁰) = 9.024 mm

δ_w = 5wL⁴/(384EI) = 5·5·1500⁴/(384·9.35 × 10¹⁰) = 3.525 mm

Add them

δ_max = 9.024 + 3.525 = 12.55 mm

Answer

δ_max = 12.55 mm

13.Column buckling

Answer: P_cr = 35.88 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·636173/3500² = 35879 N = 35.88 kN

Answer

P_cr = 35.88 kN

14.Column buckling

Answer: d = 74.02 mm

Required critical load

P_cr = n·P = 2.5·95000 = 237500 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·237500·3500²/(π³·200000))^(1/4) = 74.02 mm

Answer

d = 74.02 mm

Check: the slenderness KL/r is 189.1, so the column is slender and Euler applies.

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