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 steel bar under the axial load.

P = 30 kNd = 12 mmL = 1 mE = 200 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 = 85 kNsegment 1: Ø25 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 0.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 3 m; segment BC has diameter 20 mm and length 1 m. A torque of 200 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 200 N·md₁ = 50 mm, L₁ = 3 md₂ = 20 mm, L₂ = 1 m

τ_max = MPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 200 N·m is applied at C, 1 m from A and 1.2 m from B. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT₀ = 200 N·ma = 1 mb = 1.2 m

τ_max = MPa

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 2 kN/m from 0 to 10 mx = 3 m

V_x = kN

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 15 kN at 2.5 mw = 5 kN/m from 0 to 6 mx = 2.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 220 mm wide and 22 mm thick, an overall depth of 400 mm and a web 10 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 220 mmH = 400 mmt_f = 22 mmt_w = 10 mm

S_x = × 10³ mm³

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 60 mm wide and 20 mm high, centred on the 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 = 120 mmh = 300 mmcut-out: 60 × 20 mmhole centre: y = 180 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 30 MPa, σᵧ = −30 MPa and τₓᵧ = 20 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 = 30 MPaσ_y = −30 MPaτ_xy = 20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = −40 MPa and τₓᵧ = 40 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 140 MPaσ_y = −40 MPaτ_xy = 40 MPaσ_Y = 275 MPa

n =

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 4 kNw = 10 kN/mL = 3 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 5 mP = 25 kN at 2.5 m

M_A = kN·m

13.Column bucklingMedium

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

L = 4 mE = 70 GPa (aluminium)b × h = 40 × 50 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 110 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 110 kNn = 3L = 1.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 1.326 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·12²/4 = 113.1 mm²

σ = P/A = 30·1000/113.1 = 265.3 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 265.3/200000 = 0.001326 = 1326 µε

Elongation

δ = ε·L = 0.001326·1000 mm = 1.326 mm

Answer

δ = 1.326 mm

2.Axial stress & strain

Answer: δ = 4.257 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 173.2 MPa

σ₂ = 422.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 173.2·500/70000 = 1.237 mm

δ₂ = σ₂L₂/E₂ = 422.8·500/70000 = 3.02 mm

δ = δ₁ + δ₂ = 4.257 mm

Answer

δ = 4.257 mm

3.Torsion of shafts

Answer: τ_max = 127.3 MPa

Stress in each segment

T = 200 N·m = 200000 N·mm

τ_AB = 16·200000/(π·50³) = 8.149 MPa

τ_BC = 16·200000/(π·20³) = 127.3 MPa

Answer

τ_max = 127.3 MPa

4.Torsion of shafts

Answer: τ_max = 8.681 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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 200

T_A·a = T_B·b

T_A = 200·1.2/2.2 = 109.1 N·m

T_B = 200·1/2.2 = 90.91 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τ_max = 16·109091/(π·40³) = 8.681 MPa

Answer

τ_max = 8.681 MPa

5.Shear force & bending moment

Answer: V_x = 4 kN

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 = 2·10 = 20 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 2·10·5

R_B = 100/10 = 10 kN

ΣF_y = 0: R_A = 20 − 10 = 10 kN

Cut at x = 3 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 = 10 − 2·3 = 4 kN

Answer

V_x = 4 kN

6.Shear force & bending moment

Answer: M_x = 43.75 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·6 = 30 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 15·2.5 + 5·6·3

R_B = 127.5/6 = 21.25 kN

ΣF_y = 0: R_A = 45 − 21.25 = 23.75 kN

Cut at x = 2.5 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.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 23.75·2.5 − 5·2.5·1.25 = 43.75 kN·m

Answer

M_x = 43.75 kN·m

7.Centroid & second moment of area

Answer: S_x = 1919 × 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 = 400 − 2·22 = 356 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (220·400³ − 210·356³)/12

I_x = 3.838 × 10⁸ mm⁴

Section modulus

c = H/2 = 200 mm

S_x = I/c = 3.838 × 10⁸/200

S_x = 1.919 × 10⁶ mm³ = 1919 × 10³ mm³

Answer

S_x = 1919 × 10³ mm³

8.Centroid & second moment of area

Answer: I_x = 268.8 × 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 = (36000·150 − 1200·180)/(36000 − 1200) = 149 mm

Full plate about ȳ

120·300³/12 + 36000·1.034² = 270 × 10⁶ mm⁴

Hole about ȳ

60·20³/12 + 1200·31.03² = 1.196 × 10⁶ mm⁴

Subtract

I_x = 268.8 × 10⁶ mm⁴

Answer

I_x = 268.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 16.85 °

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(2·20, 30 − −30) = 33.69°

Angle on the element

θ_p = 16.85°

Answer

θ_p = 16.85°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 1.547

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 = 50 MPa, R = 98.49 MPa

σ₁ = 148.5 MPa, σ₂ = −48.49 MPa, σ₃ = 0

Von Mises

σ_vM = √(148.5² − 148.5·(−48.49) + (−48.49)²) = 177.8 MPa

n = 275/177.8 = 1.547

Answer

n = 1.547

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

11.Beam deflection

Answer: δ = 77.98 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.

P = 4000 N, w = 10 N/mm, L = 3000 mm

EI = 1.76 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 4000·3000³/(3·1.76 × 10¹²) = 20.45 mm

δ_w = wL⁴/(8EI) = 10·3000⁴/(8·1.76 × 10¹²) = 57.53 mm

Add them

δ = 20.45 + 57.53 = 77.98 mm

Answer

δ = 77.98 mm

12.Beam deflection

Answer: M_A = 23.44 kN·m

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = 5PL³/(48EI) → R_B = 5P/16 = 5·25/16 = 7.813 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = P·L/2 − R_B·L = 62.5 − 39.06 = 23.44 kN·m

Answer

M_A = 23.44 kN·m

13.Column buckling

Answer: P_cr = 2.879 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 = 50·40³/12 = 266667 mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·266667/8000² = 2879 N = 2.879 kN

Answer

P_cr = 2.879 kN

14.Column buckling

Answer: d = 52.61 mm

Required critical load

P_cr = n·P = 3·110000 = 330000 N

KL = 1·1500 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·330000·1500²/(π³·200000))^(1/4) = 52.61 mm

Answer

d = 52.61 mm

Check: the slenderness KL/r is 114, so the column is slender and Euler applies.

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