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 brass bar under the axial load.

P = 20 kNd = 12 mmL = 0.5 mE = 100 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø80/Ø60 mm, E = 70 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 40 mm and length 1 m; segment BC has diameter 25 mm and length 0.5 m. A torque of 750 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 750 N·md₁ = 40 mm, L₁ = 1 md₂ = 25 mm, L₂ = 0.5 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft must transmit 10 kW at 1000 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 10 kWN = 1000 rpmd_i/d_o = 0.5τ_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.

50 kN @ 4 m12 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 = 8 mP = 65 kN at 5 mw = 12 kN/m from 0 to 8 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 160 mm wide with two upright legs, 190 mm high overall. Base and legs are all 20 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 160 mmh = 190 mmt = 20 mm

ȳ = mm

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 50 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 240 mmt_f = 30 mmt_w = 14 mmh_w = 260 mmM = 50 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 90 MPa and σᵧ = 10 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 120 MPa?

σ_x = 90 MPaσ_y = 10 MPaσ₁ = 120 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −10 MPa, σᵧ = −30 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = −10 MPaσ_y = −30 MPaτ_xy = −10 MPaθ = 50°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

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

δ = 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 = 15 kN/mL = 3 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

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 = 3 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 100 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 100 kNn = 2.5L = 2 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.8842 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 = 20·1000/113.1 = 176.8 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 176.8/100000 = 0.001768 = 1768 µε

Elongation

δ = ε·L = 0.001768·500 mm = 0.8842 mm

Answer

δ = 0.8842 mm

2.Axial stress & strain

Answer: P_max = 303.9 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·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

Governing strain

ε = min(0.00075, 0.001) = 0.00075 (steel)

Largest load

P_max = 0.00075·4.053 × 10⁸ = 303949 N = 303.9 kN

Answer

P_max = 303.9 kN

3.Torsion of shafts

Answer: τ_max = 244.5 MPa

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.

Stress in each segment

T = 750 N·m = 750000 N·mm

τ_AB = 16·750000/(π·40³) = 59.68 MPa

τ_BC = 16·750000/(π·25³) = 244.5 MPa

Answer

τ_max = 244.5 MPa

4.Torsion of shafts

Answer: d_o = 23.49 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 10000/104.7 = 95.49 N·m

Outer diameter from the allowable stress

T = 95.49 N·m = 95493 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·95493/(π·40·0.9375)) = 23.49 mm

d_i = 0.5·23.49 = 11.75 mm

Answer

d_o = 23.49 mm

A solid shaft would need d = 22.99 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 = 52.67 kN, R₂ = 69.33 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 = 46.67, M_C = 81.33, M_D = 104, M_E = 114.7 kN·m, M_F = 63.33, M_G = 0

Answer

Mark E: M_E = 114.7 kN·m

6.Shear force & bending moment

Answer: M_max = 211.9 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 = 12·8 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 65·5 + 12·8·4

R_B = 709/8 = 88.63 kN

ΣF_y = 0: R_A = 161 − 88.63 = 72.38 kN

Where is V = 0?

V(5⁻) = R_A − w·5 = 12.38 kN

V(5⁺) = 12.38 − 65 = −52.63 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·5 − w·5²/2 = 72.38·5 − 12·5²/2 = 211.9 kN·m

Answer

M_max = 211.9 kN·m

It occurs at x = 5 m from A.

7.Centroid & second moment of area

Answer: ȳ = 74.6 mm

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 = 160·20 = 3200 mm², y₁ = 10 mm

two legs: A₂ = 2·20·170 = 6800 mm², y₂ = 20 + 170/2 = 105 mm

Centroid

ȳ = (3200·10 + 6800·105)/10000 = 74.6 mm

Answer

ȳ = 74.6 mm

8.Centroid & second moment of area

Answer: σ_top = 44.3 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 240·30 = 7200 mm², y_f = 260 + 30/2 = 275 mm

web: A_w = 14·260 = 3640 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (7200·275 + 3640·130)/10840 = 226.3 mm

Parallel-axis theorem

flange: 240·30³/12 + 7200·(275 − 226.3)² = 1.761 × 10⁷ mm⁴

web: 14·260³/12 + 3640·(130 − 226.3)² = 5.427 × 10⁷ mm⁴

I_x = 1.761 × 10⁷ + 5.427 × 10⁷ = 7.188 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 290 − 226.3 = 63.69 mm

σ = M·y/I = 5 × 10⁷·63.69/7.188 × 10⁷ = 44.3 MPa

Answer

σ_top = 44.3 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 57.45 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 = (90 + 10)/2 = 50 MPa

R = σ₁ − C = 120 − 50 = 70 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 40²) = 57.45 MPa

Answer

τ_xy = 57.45 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −31.58 MPa

Transformation equation

σ_θ = −20 + 10·−0.1736 + −10·0.9848 = −31.58 MPa

Answer

σ_θ = −31.58 MPa

11.Beam deflection

Answer: δ = 13.78 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 = wL⁴/(8EI) = 8·3000⁴/(8·5.88 × 10¹²) = 13.78 mm

Answer

δ = 13.78 mm

12.Beam deflection

Answer: I_req = 27.12 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 3000/360 = 8.333 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·15·3000⁴/(384·70000·8.333) = 2.712 × 10⁷ mm⁴

I_req = 27.12 × 10⁶ mm⁴

Answer

I_req = 27.12 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 284.8 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/2100² = 284752 N = 284.8 kN

Answer

P_cr = 284.8 kN

14.Column buckling

Answer: d = 61.65 mm

Required critical load

P_cr = n·P = 2.5·100000 = 250000 N

KL = 0.7·2000 = 1400 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·250000·1400²/(π³·70000))^(1/4) = 61.65 mm

Answer

d = 61.65 mm

Check: the slenderness KL/r is 90.83, so the column is slender and Euler applies.

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