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 axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 60 kNd = 30 mmL = 1.5 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

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

P = 30 kNsegment 1: Ø25 mm, 1 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel 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 = 750 N·md = 40 mmL = 1 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 1450 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 5 kWN = 1450 rpmτ_allow = 80 MPa

d = mm

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.

55 kN @ 4 m8 kN/m, 0–6 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 simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 5 mP = 20 kN at 4 mw = 10 kN/m from 0 to 5 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 30 mmhole centre: y = 45 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 150 mm plate has a rectangular cut-out 40 mm wide and 50 mm high, centred on the vertical centre line with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 150 mmcut-out: 40 × 50 mmhole centre: y = 45 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

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

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 0 MPa, σᵧ = −10 MPa and τₓᵧ = 50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σ_x = 0 MPaσ_y = −10 MPaτ_xy = 50 MPaθ = 40°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_B at the free end B.

P = 10 kNa = 1 mL = 2 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 20 kN/mL = 8 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

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 = 6 mE = 70 GPa (aluminium)b × h = 40 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 2 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 424.4 µε

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 = 60·1000/706.9 = 84.88 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 84.88/200000 = 0.0004244 = 424.4 µε

Answer

ε = 424.4 µε

2.Axial stress & strain

Answer: δ = 4.84 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·12²/4 = 113.1 mm²

σ₁ = 61.12 MPa

σ₂ = 265.3 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 61.12·1000/200000 = 0.3056 mm

δ₂ = σ₂L₂/E₂ = 265.3·2000/117000 = 4.534 mm

δ = δ₁ + δ₂ = 4.84 mm

Answer

δ = 4.84 mm

3.Torsion of shafts

Answer: φ = 2.137 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1000/(80000·251327) = 0.0373 rad

φ = 0.0373·180/π = 2.137°

Answer

φ = 2.137°

4.Torsion of shafts

Answer: d = 12.8 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 5000/151.8 = 32.93 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·32929/(π·80)) = 12.8 mm

Answer

d = 12.8 mm

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

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 = 42.33 kN, R₂ = 60.67 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 = 38.33, M_C = 68.67, M_D = 91, M_E = 105.3 kN·m, M_F = 56.67, M_G = 0

Answer

Mark E: M_E = 105.3 kN·m

6.Shear force & bending moment

Answer: M_max = 42.05 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.

UDL resultant: W = w·ℓ = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 20·4 + 10·5·2.5

R_B = 205/5 = 41 kN

ΣF_y = 0: R_A = 70 − 41 = 29 kN

Where is V = 0?

R_A − w·x₀ = 0 → x₀ = 29/10 = 2.9 m

Maximum moment

M_max = R_A·x₀ − w·x₀²/2 = 29·2.9 − 10·2.9²/2 = 42.05 kN·m

Answer

M_max = 42.05 kN·m

It occurs at x = 2.9 m from A.

7.Centroid & second moment of area

Answer: I_x = 27.42 × 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 = (15000·75 − 706.9·45)/(15000 − 706.9) = 76.48 mm

Full plate about ȳ

100·150³/12 + 15000·−1.484² = 28.16 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−31.48² = 0.7404 × 10⁶ mm⁴

Subtract

I_x = 27.42 × 10⁶ mm⁴

Answer

I_x = 27.42 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 39.8 × 10⁶ mm⁴

Centroid

ȳ = (22500·75 − 2000·45)/(22500 − 2000) = 77.93 mm

Full plate about ȳ

150·150³/12 + 22500·−2.927² = 42.38 × 10⁶ mm⁴

Hole about ȳ

40·50³/12 + 2000·−32.93² = 2.585 × 10⁶ mm⁴

Subtract

I_x = 39.8 × 10⁶ mm⁴

Answer

I_x = 39.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 25.1 °

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(2·30, 30 − −20) = 50.19°

Angle on the element

θ_p = 25.1°

Answer

θ_p = 25.1°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 45.11 MPa

Transformation equation

σ_θ = −5 + 5·0.1736 + 50·0.9848 = 45.11 MPa

Answer

σ_θ = 45.11 MPa

11.Beam deflection

Answer: δ_B = 1.812 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 = 10000 N, a = 1000 mm, L = 2000 mm

EI = 4.6 × 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) = 0.7246 mm

θ_P = Pa²/(2EI) = 0.001087 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 1.812 mm

Answer

δ_B = 1.812 mm

12.Beam deflection

Answer: I_req = 240 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 8000/360 = 22.22 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·20·8000⁴/(384·200000·22.22) = 2.4 × 10⁸ mm⁴

I_req = 240 × 10⁶ mm⁴

Answer

I_req = 240 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 8.188 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·426667/6000² = 8188 N = 8.188 kN

Answer

P_cr = 8.188 kN

14.Column buckling

Answer: σ_cr = 157.4 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 0.7·2000 = 1400 mm

KL/r = 1400/12.5 = 112

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/112² = 157.4 MPa

Answer

σ_cr = 157.4 MPa

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