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 = 105 kNd = 25 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 largest normal stress σ_max in the bar.

P = 15 kNsegment 1: Ø35 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid brass shaft 1 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 250 N·mL = 1 mG = 39 GPaφ_allow = 2°

d = mm

4.Torsion of shaftsHard

A hollow shaft must transmit 100 kW at 750 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 100 kWN = 750 rpmd_i/d_o = 0.7τ_allow = 40 MPa

d_o = 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.

60 kN @ 4 m25 kN @ 3 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 = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 60 kN at 1.5 mw = 2 kN/m from 0 to 6 mx = 1.5 m

M_x = kN·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 = 200 mmt_f = 30 mmt_w = 10 mmh_w = 200 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 40 mm wide and 20 mm high, centred on the vertical centre line with its centre 72 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmcut-out: 40 × 20 mmhole centre: y = 72 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 60 MPa, σᵧ = −50 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 60 MPaσ_y = −50 MPaτ_xy = −20 MPaθ = 25°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = 80 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σ_x = 160 MPaσ_y = 80 MPaτ_xy = 20 MPaθ = 20°

σ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 19 kNa = 1 mL = 8 mE = 70 GPa (aluminium)I = 23 × 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/250. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 4.5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 2.5 mE = 200 GPa (steel)b × h = 40 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 140 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 140 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 4.571 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·25²/4 = 490.9 mm²

σ = P/A = 105·1000/490.9 = 213.9 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 213.9/117000 = 0.001828 = 1828 µε

Elongation

δ = ε·L = 0.001828·2500 mm = 4.571 mm

Answer

δ = 4.571 mm

2.Axial stress & strain

Answer: σ_max = 74.6 MPa

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·16²/4 = 201.1 mm²

σ₁ = 15.59 MPa

σ₂ = 74.6 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 74.6 MPa

Answer

σ_max = 74.6 MPa

3.Torsion of shafts

Answer: d = 36.98 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 = 250 N·m = 250000 N·mm

L = 1 m = 1000 mm

G = 39 GPa = 39000 MPa

φ = 2·π/180 = 0.03491 rad

J = TL/(Gφ) = 250000·1000/(39000·0.03491) = 183640 mm⁴

Diameter

d = (32J/π)^1/4 = 36.98 mm

Answer

d = 36.98 mm

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

4.Torsion of shafts

Answer: d_o = 59.75 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 100000/78.54 = 1273 N·m

Outer diameter from the allowable stress

T = 1273 N·m = 1.273 × 10⁶ N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1.273 × 10⁶/(π·40·0.7599)) = 59.75 mm

d_i = 0.7·59.75 = 41.83 mm

Answer

d_o = 59.75 mm

A solid shaft would need d = 54.53 mm; the hollow one is a little larger but much lighter.

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 = 32.5 kN, R₂ = 52.5 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 = 32.5, M_C = 65, M_D = 97.5, M_E = 105 kN·m, M_F = 52.5, M_G = 0

Answer

Mark E: M_E = 105 kN·m

6.Shear force & bending moment

Answer: M_x = 74.25 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 = 2·6 = 12 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 60·1.5 + 2·6·3

R_B = 126/6 = 21 kN

ΣF_y = 0: R_A = 72 − 21 = 51 kN

Cut at x = 1.5 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 = 51·1.5 − 2·1.5·0.75 = 74.25 kN·m

Answer

M_x = 74.25 kN·m

7.Centroid & second moment of area

Answer: I_x = 26.95 × 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 = 200·30 = 6000 mm², y_f = 200 + 30/2 = 215 mm

web: A_w = 10·200 = 2000 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·215 + 2000·100)/8000 = 186.3 mm

Parallel-axis theorem

flange: 200·30³/12 + 6000·(215 − 186.3)² = 5.409 × 10⁶ mm⁴

web: 10·200³/12 + 2000·(100 − 186.3)² = 2.154 × 10⁷ mm⁴

I_x = 5.409 × 10⁶ + 2.154 × 10⁷ = 2.695 × 10⁷ mm⁴ = 26.95 × 10⁶ mm⁴

Answer

I_x = 26.95 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 136.3 × 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 − 800·72)/(28800 − 800) = 121.4 mm

Full plate about ȳ

120·240³/12 + 28800·−1.371² = 138.3 × 10⁶ mm⁴

Hole about ȳ

40·20³/12 + 800·−49.37² = 1.977 × 10⁶ mm⁴

Subtract

I_x = 136.3 × 10⁶ mm⁴

Answer

I_x = 136.3 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = −15.03 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.

Transformation equation for the y′ face

σ_y′ = 5 − 55·0.6428 − (−20)·0.766 = −15.03 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 25.03 MPa, σ_x′ + σ_y′ = 10 = σ_x + σ_y

Answer

σ_y′ = −15.03 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 163.5 MPa

Transformation equation

σ_θ = 120 + 40·0.766 + 20·0.6428 = 163.5 MPa

Answer

σ_θ = 163.5 MPa

11.Beam deflection

Answer: δ_P = 24.09 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 = 19000 N, a = 1000 mm, b = 7000 mm, L = 8000 mm

EI = 1.61 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 19000·1000²·7000²/(3·1.61 × 10¹²·8000) = 24.09 mm

Answer

δ_P = 24.09 mm

12.Beam deflection

Answer: h_req = 200.7 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 = 2.5 N/mm, L = 4500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 4500/250 = 18 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·4500⁴/(384·11000·18) = 6.742 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·6.742 × 10⁷/100)^1/3 = 200.7 mm

Answer

h_req = 200.7 mm

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

13.Column buckling

Answer: P_cr = 539 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 = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·426667/1250² = 539012 N = 539 kN

Answer

P_cr = 539 kN

14.Column buckling

Answer: d = 106.8 mm

Required critical load

P_cr = n·P = 2.5·140000 = 350000 N

KL = 2·3000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·350000·6000²/(π³·200000))^(1/4) = 106.8 mm

Answer

d = 106.8 mm

Check: the slenderness KL/r is 224.7, so the column is slender and Euler applies.

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