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 = 30 kNd = 10 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 total elongation δ.

P = 70 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø20 mm, 2 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid brass shaft 1.5 m long carries a torque of 500 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 500 N·mL = 1.5 mG = 39 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1000 rpmd_o = 40 mmd_i = 24 mm

τ_max = MPa

5.Shear force & bending momentMedium

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.

45 kN @ 5 m35 kN @ 4 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)

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 m40 kN @ 12 m45 kN @ 6 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 120 mmt_f = 20 mmt_w = 14 mmh_w = 180 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 200 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 140 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: 80 × 20 mmhole centre: y = 140 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 0 MPa and σᵧ = −50 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 60 MPa?

σ_x = 0 MPaσ_y = −50 MPaσ₁ = 60 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries an axial compressive force of 200 kN together with a torque of 1000 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 40 mmF = −200 kNT = 1000 N·m

σ₁ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1.5 m from A. Find the deflection δ_B at the free end B.

P = 25 kNa = 1.5 mL = 4 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 8 mw = 15 kN/m

R_B = kN

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 solid round column of length L made of steel is fixed at both ends. It must carry 15 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 15 kNn = 2L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 8.162 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·10²/4 = 78.54 mm²

σ = P/A = 30·1000/78.54 = 382 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 382/117000 = 0.003265 = 3265 µε

Elongation

δ = ε·L = 0.003265·2500 mm = 8.162 mm

Answer

δ = 8.162 mm

2.Axial stress & strain

Answer: δ = 7.299 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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 72.76 MPa

σ₂ = 222.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 72.76·1500/117000 = 0.9328 mm

δ₂ = σ₂L₂/E₂ = 222.8·2000/70000 = 6.366 mm

δ = δ₁ + δ₂ = 7.299 mm

Answer

δ = 7.299 mm

3.Torsion of shafts

Answer: d = 68.83 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 = 500 N·m = 500000 N·mm

L = 1.5 m = 1500 mm

G = 39 GPa = 39000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 500000·1500/(39000·0.008727) = 2.204 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 68.83 mm

Answer

d = 68.83 mm

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

4.Torsion of shafts

Answer: τ_max = 87.31 MPa

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 100000/104.7 = 954.9 N·m

Polar second moment of area

J = π(40⁴ − 24⁴)/32 = 218755 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·20/218755 = 87.31 MPa

Answer

τ_max = 87.31 MPa

5.Shear force & bending moment

Answer: E (x = 4 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 = 19.17 kN, R₂ = 60.83 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 = 19.17, M_C = 38.33, M_D = 57.5, M_E = 76.67 kN·m, M_F = 60.83, M_G = 0

Answer

Mark E: M_E = 76.67 kN·m

6.Shear force & bending moment

Answer: E (x = 8 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 15 kN, R₂ = 90 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 = −40, M_C = −50, M_D = −60, M_E = −160 kN·m, M_F = −80, M_G = 0

Answer

Mark E: M_E = −160 kN·m

7.Centroid & second moment of area

Answer: I_x = 19.18 × 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 two rectangles

Measure every y from the bottom edge.

flange: A_f = 120·20 = 2400 mm², y_f = 180 + 20/2 = 190 mm

web: A_w = 14·180 = 2520 mm², y_w = 180/2 = 90 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·190 + 2520·90)/4920 = 138.8 mm

Parallel-axis theorem

flange: 120·20³/12 + 2400·(190 − 138.8)² = 6.376 × 10⁶ mm⁴

web: 14·180³/12 + 2520·(90 − 138.8)² = 1.28 × 10⁷ mm⁴

I_x = 6.376 × 10⁶ + 1.28 × 10⁷ = 1.918 × 10⁷ mm⁴ = 19.18 × 10⁶ mm⁴

Answer

I_x = 19.18 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 77.2 × 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 = (24000·100 − 1600·140)/(24000 − 1600) = 97.14 mm

Full plate about ȳ

120·200³/12 + 24000·2.857² = 80.2 × 10⁶ mm⁴

Hole about ȳ

80·20³/12 + 1600·42.86² = 2.992 × 10⁶ mm⁴

Subtract

I_x = 77.2 × 10⁶ mm⁴

Answer

I_x = 77.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 81.24 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 + (−50))/2 = −25 MPa

R = σ₁ − C = 60 − (−25) = 85 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(85² − 25²) = 81.24 MPa

Answer

τ_xy = 81.24 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 32.96 MPa

Stresses at the surface

σ = F/A = −200000/(π·40²/4) = −200000/1257 = −159.2 MPa

τ = 16T/(πd³) = 16·1000000/(π·40³) = 79.58 MPa

Mohr’s circle

C = σ/2 = −79.58 MPa

R = √((−79.58)² + 79.58²) = 112.5 MPa

σ₁ = C + R = 32.96 MPa

σ₂ = C − R = −192.1 MPa

Answer

σ₁ = 32.96 MPa

11.Beam deflection

Answer: δ_B = 57.9 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 = 25000 N, a = 1500 mm, L = 4000 mm

EI = 1.7 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 16.54 mm

θ_P = Pa²/(2EI) = 0.01654 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 57.9 mm

Answer

δ_B = 57.9 mm

12.Beam deflection

Answer: R_B = 45 kN

Remove the prop

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

δ_load = wL⁴/(8EI)

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) = wL⁴/(8EI) → R_B = 3wL/8 = 3·15·8/8 = 45 kN

Answer

R_B = 45 kN

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: d = 33.36 mm

Required critical load

P_cr = n·P = 2·15000 = 30000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·30000·2000²/(π³·200000))^(1/4) = 33.36 mm

Answer

d = 33.36 mm

Check: the slenderness KL/r is 239.8, so the column is slender and Euler applies.

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