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 copper bar under the axial load.

P = 110 kNd = 30 mmL = 2.5 mE = 117 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 = 10 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 2 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 25 mm and length 2.5 m; segment BC has diameter 20 mm and length 1.5 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 250 N·md₁ = 25 mm, L₁ = 2.5 md₂ = 20 mm, L₂ = 1.5 m

τ_max = MPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 50 mm is fixed at A. A torque of 400 N·m acts at B, 0.5 m from A, and a torque of 400 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τ_max in the shaft.

d = 50 mmT_B = 400 N·mT_C = 400 N·mAB = 0.5 mBC = 2.5 m

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 m from A. Find the bending moment M_x at the section x = 0.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 2 mP = 20 kN at 2 mw = 12 kN/m from 0 to 2 mx = 0.5 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 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 5.5 mw = 8 kN/m from 0 to 5.5 msupports at x = 0 and x = 4 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 300 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 180 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 300 mmd = 60 mmhole centre: y = 180 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 200 mm plate has a rectangular cut-out 60 mm wide and 20 mm high, centred on the 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 = 120 mmh = 200 mmcut-out: 60 × 20 mmhole centre: y = 60 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 0 MPa and σᵧ = 30 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 55 MPa?

σ_x = 0 MPaσ_y = 30 MPaσ₁ = 55 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 10 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1500 mmt = 10 mmp = 0.8 MPaα = 35°

σ_n = 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 = 20 kN/mL = 2.5 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ = 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 = 2 kN/mL = 4.5 mE = 11 GPa (timber)b = 150 mm

h_req = mm

13.Column bucklingMedium

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

L = 6 mE = 200 GPa (steel)d = 25 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 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 5 kNn = 3L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 3.325 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 = 110·1000/706.9 = 155.6 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 155.6/117000 = 0.00133 = 1330 µε

Elongation

δ = ε·L = 0.00133·2500 mm = 3.325 mm

Answer

δ = 3.325 mm

2.Axial stress & strain

Answer: σ_max = 31.83 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 = 10.39 MPa

σ₂ = 31.83 MPa

Largest stress

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

σ_max = 31.83 MPa

Answer

σ_max = 31.83 MPa

3.Torsion of shafts

Answer: τ_max = 159.2 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 = 250 N·m = 250000 N·mm

τ_AB = 16·250000/(π·25³) = 81.49 MPa

τ_BC = 16·250000/(π·20³) = 159.2 MPa

Answer

τ_max = 159.2 MPa

4.Torsion of shafts

Answer: τ_max = 32.59 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 400 N·m

T_AB = T_C + T_B = 400 + 400 = 800 N·m

Maximum shear stress

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

|T|_max = 800 N·m

τ_max = 16·800000/(π·50³) = 32.59 MPa

Answer

τ_max = 32.59 MPa

5.Shear force & bending moment

Answer: M_x = −43.5 kN·m

Keep the free-end part

P: −20·1.5 = −30 kN·m

w: −12·1.5²/2 = −13.5 kN·m

M_x = −30 − 13.5 = −43.5 kN·m

Answer

M_x = −43.5 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_max = 11.82 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 8·5.5 = 44 kN, acting at x̄ = 2.75 m

ΣM_A = 0: R_B·4 = 8·5.5·2.75

R_B = 121/4 = 30.25 kN

ΣF_y = 0: R_A = 44 − 30.25 = 13.75 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 13.75/8 = 1.719 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 13.75²/(2·8) = 11.82 kN·m

Answer

M_max = 11.82 kN·m

7.Centroid & second moment of area

Answer: I_x = 446.7 × 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 = (60000·150 − 2827·180)/(60000 − 2827) = 148.5 mm

Full plate about ȳ

200·300³/12 + 60000·1.484² = 450.1 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·31.48² = 3.439 × 10⁶ mm⁴

Subtract

I_x = 446.7 × 10⁶ mm⁴

Answer

I_x = 446.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 77.94 × 10⁶ mm⁴

Centroid

ȳ = (24000·100 − 1200·60)/(24000 − 1200) = 102.1 mm

Full plate about ȳ

120·200³/12 + 24000·−2.105² = 80.11 × 10⁶ mm⁴

Hole about ȳ

60·20³/12 + 1200·−42.11² = 2.167 × 10⁶ mm⁴

Subtract

I_x = 77.94 × 10⁶ mm⁴

Answer

I_x = 77.94 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 37.08 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 = (0 + 30)/2 = 15 MPa

R = σ₁ − C = 55 − 15 = 40 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(40² − 15²) = 37.08 MPa

Answer

τ_xy = 37.08 MPa

10.Mohr’s circle & stress transformation

Answer: σ_n = 50.13 MPa

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 0.8·750/10 = 60 MPa

σ_a = pr/(2t) = 30 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 30·sin²35° + 60·cos²35° = 50.13 MPa

Answer

σ_n = 50.13 MPa

11.Beam deflection

Answer: δ = 73.98 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 = 20 N/mm, L = 2500 mm

EI = 1.32 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 20·2500⁴/(8·1.32 × 10¹²) = 73.98 mm

Answer

δ = 73.98 mm

12.Beam deflection

Answer: h_req = 183.8 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 = 2 N/mm, L = 4500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 4500/360 = 12.5 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2·4500⁴/(384·11000·12.5) = 7.766 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·7.766 × 10⁷/150)^1/3 = 183.8 mm

Answer

h_req = 183.8 mm

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

13.Column buckling

Answer: P_cr = 4.205 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/3000² = 4205 N = 4.205 kN

Answer

P_cr = 4.205 kN

14.Column buckling

Answer: d = 39.67 mm

Required critical load

P_cr = n·P = 3·5000 = 15000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·4000²/(π³·200000))^(1/4) = 39.67 mm

Answer

d = 39.67 mm

Check: the slenderness KL/r is 403.3, so the column is slender and Euler applies.

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