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 = 115 kNd = 12 mmL = 3 mE = 117 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 80 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid brass shaft 0.5 m long carries a torque of 100 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 100 N·mL = 0.5 mG = 39 GPaφ_allow = 1°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 300 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 10 kWN = 300 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 8 kN/m from 0 to 4 mx = 2 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 9 mw = 3 kN/m from 0 to 9 msupports at x = 0 and x = 8 m

x_c = m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 120 mm wide with two upright legs, 160 mm high overall. Base and legs are all 14 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 120 mmh = 160 mmt = 14 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 25 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 160 mmt_f = 25 mmt_w = 16 mmh_w = 260 mmM = 25 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa and σᵧ = 40 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 185 MPa?

σ_x = 130 MPaσ_y = 40 MPaσ₁ = 185 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −60 MPa, σᵧ = −80 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σ_x = −60 MPaσ_y = −80 MPaτ_xy = −30 MPaθ = 35°

σ_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 = 8 kN/mL = 3 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1 m from support B. Find the deflection δ_mid at midspan.

P = 2 kNb = 1 mL = 6 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴

δ_mid = mm

13.Column bucklingMedium

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 200 GPa (steel)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 25 kNn = 2L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 26.07 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·12²/4 = 113.1 mm²

σ = P/A = 115·1000/113.1 = 1017 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1017/117000 = 0.008691 = 8691 µε

Elongation

δ = ε·L = 0.008691·3000 mm = 26.07 mm

Answer

δ = 26.07 mm

2.Axial stress & strain

Answer: P_max = 316.3 kN

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·30²/4 = 706.9 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·4006 = 4.218 × 10⁸ N

Governing strain

ε = min(0.00075, 0.001143) = 0.00075 (steel)

Largest load

P_max = 0.00075·4.218 × 10⁸ = 316319 N = 316.3 kN

Answer

P_max = 316.3 kN

3.Torsion of shafts

Answer: d = 29.41 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 = 100 N·m = 100000 N·mm

L = 0.5 m = 500 mm

G = 39 GPa = 39000 MPa

φ = 1·π/180 = 0.01745 rad

J = TL/(Gφ) = 100000·500/(39000·0.01745) = 73456 mm⁴

Diameter

d = (32J/π)^1/4 = 29.41 mm

Answer

d = 29.41 mm

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

4.Torsion of shafts

Answer: d = 31.89 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 10000/31.42 = 318.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·318310/(π·50)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Shear force & bending moment

Answer: M_x = 16 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·4 = 32 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 8·4·2

R_B = 64/4 = 16 kN

ΣF_y = 0: R_A = 32 − 16 = 16 kN

Cut at x = 2 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 = 16·2 − 8·2·1 = 16 kN·m

Answer

M_x = 16 kN·m

6.Shear force & bending moment

Answer: x_c = 7.875 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 3·9 = 27 kN, acting at x̄ = 4.5 m

ΣM_A = 0: R_B·8 = 3·9·4.5

R_B = 121.5/8 = 15.19 kN

ΣF_y = 0: R_A = 27 − 15.19 = 11.81 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·11.81/3 = 7.875 m

Answer

x_c = 7.875 m

7.Centroid & second moment of area

Answer: I_x = 14.91 × 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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 120·14 = 1680 mm², y₁ = 7 mm

two legs: A₂ = 2·14·146 = 4088 mm², y₂ = 14 + 146/2 = 87 mm

Centroid

ȳ = (1680·7 + 4088·87)/5768 = 63.7 mm

Parallel-axis theorem

base: 120·14³/12 + 1680·(7 − 63.7)² = 5.428 × 10⁶ mm⁴

two legs: 2·14·146³/12 + 4088·(87 − 63.7)² = 9.481 × 10⁶ mm⁴

I_x = 5.428 × 10⁶ + 9.481 × 10⁶

I_x = 1.491 × 10⁷ mm⁴ = 14.91 × 10⁶ mm⁴

Answer

I_x = 14.91 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 76.81 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 160·25 = 4000 mm², y_f = 260 + 25/2 = 272.5 mm

web: A_w = 16·260 = 4160 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·272.5 + 4160·130)/8160 = 199.9 mm

Parallel-axis theorem

flange: 160·25³/12 + 4000·(272.5 − 199.9)² = 2.132 × 10⁷ mm⁴

web: 16·260³/12 + 4160·(130 − 199.9)² = 4.373 × 10⁷ mm⁴

I_x = 2.132 × 10⁷ + 4.373 × 10⁷ = 6.505 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 199.9 mm

σ = M·y/I = 2.5 × 10⁷·199.9/6.505 × 10⁷ = 76.81 MPa

Answer

σ_bottom = 76.81 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 89.3 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 = (130 + 40)/2 = 85 MPa

R = σ₁ − C = 185 − 85 = 100 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(100² − 45²) = 89.3 MPa

Answer

τ_xy = 89.3 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −94.77 MPa

Transformation equation

σ_θ = −70 + 10·0.342 + −30·0.9397 = −94.77 MPa

Answer

σ_θ = −94.77 MPa

11.Beam deflection

Answer: δ_max = 1.435 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 = 8 N/mm, L = 3000 mm

EI = 5.88 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·3000⁴/(384·5.88 × 10¹²) = 1.435 mm

Answer

δ_max = 1.435 mm

12.Beam deflection

Answer: δ_mid = 10.94 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 = 2000 N, b = 1000 mm, L = 6000 mm

EI = 3.96 × 10¹¹ N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 2000·1000·(3·6000² − 4·1000²)/(48·3.96 × 10¹¹) = 10.94 mm

Answer

δ_mid = 10.94 mm

13.Column buckling

Answer: P_cr = 45.32 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·180000/2800² = 45320 N = 45.32 kN

Answer

P_cr = 45.32 kN

14.Column buckling

Answer: d = 59.93 mm

Required critical load

P_cr = n·P = 2·25000 = 50000 N

KL = 1·5000 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·5000²/(π³·200000))^(1/4) = 59.93 mm

Answer

d = 59.93 mm

Check: the slenderness KL/r is 333.7, so the column is slender and Euler applies.

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