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 = 50 kNd = 20 mmL = 2 mE = 117 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: copper, Ø30 mm, 400 mm, E = 120 GPaP = 80 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 100 N·md = 30 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1450 rpmd_o = 60 mmd_i = 48 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 45 kN·m at x = 2 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 4 mM₀ = 45 kN·m clockwise, at 2 m

M_x = kN·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.

60 kN @ 0 m10 kN @ 12 m40 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

A 120 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 96 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmd = 50 mmhole centre: y = 96 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 120 × 22 mm, a web 150 mm high and 8 mm thick, and a bottom flange 190 × 20 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 120 × 22 mmweb: 8 × 150 mmbottom flange: 190 × 20 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 130 MPaσ_y = −60 MPaτ_xy = 20 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −40 MPa, σᵧ = −40 MPa and τₓᵧ = 60 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −40 MPaσ_y = −40 MPaτ_xy = 60 MPaθ = 40°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 3 m from support A. Find the deflection δ_P under the load.

P = 22 kNa = 3 mL = 5 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 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 = 1 kN/mL = 3.5 mE = 11 GPa (timber)b = 100 mm

h_req = 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)b × h = 20 × 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 2 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 2.721 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·20²/4 = 314.2 mm²

σ = P/A = 50·1000/314.2 = 159.2 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 159.2/117000 = 0.00136 = 1360 µε

Elongation

δ = ε·L = 0.00136·2000 mm = 2.721 mm

Answer

δ = 2.721 mm

2.Axial stress & strain

Answer: σ₂ = 42.44 MPa

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.

Stiffnesses

k₁ = 200000·706.9/400 = 353429 N/mm

k₂ = 120000·706.9/400 = 212058 N/mm

Share of segment 2

F₂ = 80000·212058/565487 = 30000 N

Stress

σ₂ = 30000/706.9 = 42.44 MPa (compression)

Answer

σ₂ = 42.44 MPa

3.Torsion of shafts

Answer: φ = 5.542 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·2000/(26000·79522) = 0.09673 rad

φ = 0.09673·180/π = 5.542°

Answer

φ = 5.542°

4.Torsion of shafts

Answer: τ_max = 26.3 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(60⁴ − 48⁴)/32 = 751193 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 658572·30/751193 = 26.3 MPa

Answer

τ_max = 26.3 MPa

5.Shear force & bending moment

Answer: M_x = 22.5 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 45/4 = 11.25 kN (upward)

R_A = −R_B = −11.25 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −11.25·2 + 45 = 22.5 kN·m

Answer

M_x = 22.5 kN·m

6.Shear force & bending moment

Answer: B (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 = 92.5 kN, R₂ = 17.5 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 = −120 kN·m, M_C = −55, M_D = 10, M_E = −5, M_F = −20, M_G = 0

Answer

Mark B: M_B = −120 kN·m

7.Centroid & second moment of area

Answer: I_x = 136.7 × 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 = (28800·120 − 1963·96)/(28800 − 1963) = 121.8 mm

Full plate about ȳ

120·240³/12 + 28800·−1.756² = 138.3 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·−25.76² = 1.609 × 10⁶ mm⁴

Subtract

I_x = 136.7 × 10⁶ mm⁴

Answer

I_x = 136.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 48.26 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 2640 mm², y = 181 mm

web: A = 1200 mm², y = 95 mm

bottom flange: A = 3800 mm², y = 10 mm

Centroid

ȳ = (2640·181 + 1200·95 + 3800·10)/7640 = 82.44 mm

Parallel-axis theorem

top flange: 120·22³/12 + 2640·98.56² = 2.575 × 10⁷ mm⁴

web: 8·150³/12 + 1200·12.56² = 2.439 × 10⁶ mm⁴

bottom flange: 190·20³/12 + 3800·72.44² = 2.007 × 10⁷ mm⁴

I_x = 2.575 × 10⁷ + 2.439 × 10⁶ + 2.007 × 10⁷

I_x = 4.826 × 10⁷ mm⁴ = 48.26 × 10⁶ mm⁴

Answer

I_x = 48.26 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 97.08 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.

Radius of Mohr’s circle

τ_max = R = √(((130 − −60)/2)² + 20²) = 97.08 MPa

Answer

τ_max = 97.08 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 10.42 MPa

Transformation equation

τ_θ = −0·0.9848 + 60·0.1736 = 10.42 MPa

Answer

τ_θ = 10.42 MPa

11.Beam deflection

Answer: δ_P = 1.65 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 = 22000 N, a = 3000 mm, b = 2000 mm, L = 5000 mm

EI = 3.2 × 10¹³ N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 22000·3000²·2000²/(3·3.2 × 10¹³·5000) = 1.65 mm

Answer

δ_P = 1.65 mm

12.Beam deflection

Answer: h_req = 129.9 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 = 1 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 3500/360 = 9.722 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1·3500⁴/(384·11000·9.722) = 1.827 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.827 × 10⁷/100)^1/3 = 129.9 mm

Answer

h_req = 129.9 mm

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

13.Column buckling

Answer: P_cr = 8.355 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·53333/2100² = 8355 N = 8.355 kN

Answer

P_cr = 8.355 kN

14.Column buckling

Answer: σ_cr = 6.94 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 2·2000 = 4000 mm

KL/r = 4000/7.5 = 533.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/533.3² = 6.94 MPa

Answer

σ_cr = 6.94 MPa

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