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 copper bar, in microstrain (µε = 10⁻⁶).

P = 115 kNd = 30 mmL = 1.5 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 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 = 250 MPatube: Ø100/Ø70 mm, E = 110 GPa, σ_allow = 80 MPa

P_max = kN

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 400 mm rotate 3.55° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 15 mmL = 400 mmφ = 3.55°

G = GPa

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 300 rpmd_o = 50 mmd_i = 25 mm

τ_max = MPa

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.

20 kN @ 6 m10 kN/m, 4–8 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)

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.

35 kN @ 0 m25 kN @ 12 m40 kN @ 4 m6 kN/m, 2–10 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 symmetric I-section has flanges 150 mm wide and 10 mm thick, an overall depth of 300 mm and a web 10 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 150 mmH = 300 mmt_f = 10 mmt_w = 10 mm

S_x = × 10³ mm³

8.Centroid & second moment of areaHard

A welded I-section has a top flange 130 × 24 mm, a web 290 mm high and 8 mm thick, and a bottom flange 270 × 14 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 130 × 24 mmweb: 8 × 290 mmbottom flange: 270 × 14 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −10 MPa, σᵧ = 30 MPa and τₓᵧ = −20 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 = −10 MPaσ_y = 30 MPaτ_xy = −20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −40 MPa and τₓᵧ = 40 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = −70 MPaσ_y = −40 MPaτ_xy = 40 MPa

σ_vM = 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 = 6 kN/mL = 4 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ_max = 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 = 2 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1391 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 = 115·1000/706.9 = 162.7 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 162.7/117000 = 0.001391 = 1391 µε

Answer

ε = 1391 µε

2.Axial stress & strain

Answer: P_max = 423.3 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 + 110·10³·4006 = 5.82 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0007273) = 0.0007273 (bronze)

Largest load

P_max = 0.0007273·5.82 × 10⁸ = 423258 N = 423.3 kN

Answer

P_max = 423.3 kN

3.Torsion of shafts

Answer: G = 38.97 GPa

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 moment and angle

J = π·15⁴/32 = 4970 mm⁴

φ = 3.55·π/180 = 0.06196 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·400/(4970·0.06196) = 38968 MPa

G = 38.97 GPa

Answer

G = 38.97 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: τ_max = 27.67 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 20000/31.42 = 636.6 N·m

Polar second moment of area

J = π(50⁴ − 25⁴)/32 = 575243 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·25/575243 = 27.67 MPa

Answer

τ_max = 27.67 MPa

5.Shear force & bending moment

Answer: D (x = 6 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 = 30 kN, R₂ = 30 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 = 60, M_C = 120, M_D = 160 kN·m, M_E = 120, M_F = 60, M_G = 0

Answer

Mark D: M_D = 160 kN·m

6.Shear force & bending moment

Answer: B (x = 2 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 91.5 kN, R₂ = 56.5 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 = −70 kN·m, M_C = 31, M_D = 28, M_E = 1, M_F = −50, M_G = 0

Answer

Mark B: M_B = −70 kN·m

7.Centroid & second moment of area

Answer: S_x = 542.6 × 10³ mm³

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 300 − 2·10 = 280 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (150·300³ − 140·280³)/12

I_x = 8.139 × 10⁷ mm⁴

Section modulus

c = H/2 = 150 mm

S_x = I/c = 8.139 × 10⁷/150

S_x = 542622 mm³ = 542.6 × 10³ mm³

Answer

S_x = 542.6 × 10³ mm³

8.Centroid & second moment of area

Answer: I_x = 179.9 × 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 = 3120 mm², y = 316 mm

web: A = 2320 mm², y = 159 mm

bottom flange: A = 3780 mm², y = 7 mm

Centroid

ȳ = (3120·316 + 2320·159 + 3780·7)/9220 = 149.8 mm

Parallel-axis theorem

top flange: 130·24³/12 + 3120·166.2² = 8.632 × 10⁷ mm⁴

web: 8·290³/12 + 2320·9.189² = 1.646 × 10⁷ mm⁴

bottom flange: 270·14³/12 + 3780·142.8² = 7.716 × 10⁷ mm⁴

I_x = 8.632 × 10⁷ + 1.646 × 10⁷ + 7.716 × 10⁷

I_x = 1.799 × 10⁸ mm⁴ = 179.9 × 10⁶ mm⁴

Answer

I_x = 179.9 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = −67.5 °

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·−20, −10 − 30) = −135°

Angle on the element

θ_p = −67.5°

Answer

θ_p = −67.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_vM = 92.2 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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = −55 MPa, R = 42.72 MPa

σ₁ = −12.28 MPa, σ₂ = −97.72 MPa, σ₃ = 0

Von Mises

σ_vM = √((−12.28)² − (−12.28)·(−97.72) + (−97.72)²) = 92.2 MPa

Answer

σ_vM = 92.2 MPa

11.Beam deflection

Answer: δ_max = 1.19 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 = 6 N/mm, L = 4000 mm

EI = 1.68 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·6·4000⁴/(384·1.68 × 10¹³) = 1.19 mm

Answer

δ_max = 1.19 mm

12.Beam deflection

Answer: I_req = 2.813 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 2000/360 = 5.556 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·15·2000⁴/(384·200000·5.556) = 2.813 × 10⁶ mm⁴

I_req = 2.813 × 10⁶ mm⁴

Answer

I_req = 2.813 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 50.23 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 = 1·5000 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/5000² = 50230 N = 50.23 kN

Answer

P_cr = 50.23 kN

14.Column buckling

Answer: σ_cr = 69.09 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·1000 = 1000 mm

KL/r = 1000/10 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

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