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

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa). Both are 300 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 70 GPaL = 300 mm

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 200 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 200 mm, E = 110 GPaP = 50 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid aluminium shaft 2.5 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 1000 N·mL = 2.5 mG = 26 GPaφ_allow = 1°

d = mm

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 100 N·m acts at B, 1.5 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τ_max in the shaft.

d = 60 mmT_B = 100 N·mT_C = 750 N·mAB = 1.5 mBC = 0.8 m

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.

20 kN @ 1 m5 kN/m, 0–4 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

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 = 8 mP = 50 kN at 4 mw = 10 kN/m from 0 to 8 mx = 4 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 150 mm × 300 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 210 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 300 mmd = 50 mmhole centre: y = 210 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 100 mm wide and 16 mm thick, an overall depth of 420 mm and a web 8 mm thick. It carries a bending moment of 60 kN·m about the axis x. Find the size of the maximum bending stress σ_max in the section.

B = 100 mmH = 420 mmt_f = 16 mmt_w = 8 mmM = 60 kN·m

σ_max = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 20 MPa, σᵧ = 50 MPa and τₓᵧ = 40 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 = 20 MPaσ_y = 50 MPaτ_xy = 40 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 20 MPa, σᵧ = −10 MPa and τₓᵧ = 10 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 20 MPaσ_y = −10 MPaτ_xy = 10 MPaθ = -45°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 6 kN/mL = 6 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the maximum deflection δ_max of the beam.

P = 21 kNb = 1.5 mL = 6 mE = 70 GPa (aluminium)I = 36 × 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 = 3 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 105 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 105 kNn = 2L = 3 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.2221 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·2199 = 4.053 × 10⁸ N

Shortening

δ = 300000·300/4.053 × 10⁸ = 0.2221 mm

Answer

δ = 0.2221 mm

2.Axial stress & strain

Answer: σ₂ = 21.33 MPa

Stiffnesses

k₁ = 200000·314.2/300 = 209440 N/mm

k₂ = 110000·1963/200 = 1.08 × 10⁶ N/mm

Share of segment 2

F₂ = 50000·1.08 × 10⁶/1.289 × 10⁶ = 41878 N

Stress

σ₂ = 41878/1963 = 21.33 MPa (compression)

Answer

σ₂ = 21.33 MPa

3.Torsion of shafts

Answer: d = 86.55 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 = 1000 N·m = 1 × 10⁶ N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = 1·π/180 = 0.01745 rad

J = TL/(Gφ) = 1 × 10⁶·2500/(26000·0.01745) = 5.509 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 86.55 mm

Answer

d = 86.55 mm

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

4.Torsion of shafts

Answer: τ_max = 17.68 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 750 N·m

T_AB = T_C − T_B = 750 − 100 = 650 N·m

Maximum shear stress

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

|T|_max = 750 N·m

τ_max = 16·750000/(π·60³) = 17.68 MPa

Answer

τ_max = 17.68 MPa

5.Shear force & bending moment

Answer: C (x = 2 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 = 30 kN, R₂ = 10 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 = 27.5, M_C = 30 kN·m, M_D = 27.5, M_E = 20, M_F = 10, M_G = 0

Answer

Mark C: M_C = 30 kN·m

6.Shear force & bending moment

Answer: M_x = 180 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 = 10·8 = 80 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 50·4 + 10·8·4

R_B = 520/8 = 65 kN

ΣF_y = 0: R_A = 130 − 65 = 65 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 = 65·4 − 10·4·2 = 180 kN·m

Answer

M_x = 180 kN·m

7.Centroid & second moment of area

Answer: I_x = 329.8 × 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 = (45000·150 − 1963·210)/(45000 − 1963) = 147.3 mm

Full plate about ȳ

150·300³/12 + 45000·2.737² = 337.8 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·62.74² = 8.035 × 10⁶ mm⁴

Subtract

I_x = 329.8 × 10⁶ mm⁴

Answer

I_x = 329.8 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_max = 74.3 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.

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 = 420 − 2·16 = 388 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (100·420³ − 92·388³)/12

I_x = 1.696 × 10⁸ mm⁴

Bending stress

y = 210 mm

σ = M·y/I = 6 × 10⁷·210/1.696 × 10⁸ = 74.3 MPa

Answer

σ_max = 74.3 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 55.28 °

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(2·40, 20 − 50) = 110.6°

Angle on the element

θ_p = 55.28°

Answer

θ_p = 55.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 15 MPa

Transformation equation

τ_θ = −15·−1 + 10·0 = 15 MPa

Answer

τ_θ = 15 MPa

11.Beam deflection

Answer: δ_max = 8.882 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 = 6 N/mm, L = 6000 mm

EI = 1.14 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·6·6000⁴/(384·1.14 × 10¹³) = 8.882 mm

Answer

δ_max = 8.882 mm

12.Beam deflection

Answer: δ_max = 26.2 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 = 21000 N, b = 1500 mm, L = 6000 mm

EI = 2.52 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((6000² − 1500²)/3) = 3354 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 26.2 mm

Answer

δ_max = 26.2 mm

13.Column buckling

Answer: P_cr = 19.69 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/2100² = 19687 N = 19.69 kN

Answer

P_cr = 19.69 kN

14.Column buckling

Answer: d = 61.1 mm

Required critical load

P_cr = n·P = 2·105000 = 210000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·210000·1500²/(π³·70000))^(1/4) = 61.1 mm

Answer

d = 61.1 mm

Check: the slenderness KL/r is 98.21, so the column is slender and Euler applies.

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