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 steel bar under the axial load.

P = 40 kNd = 10 mmL = 3 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 70 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/Ø50 mm, E = 70 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 80 mm, inner diameter 60 mm, length 2.5 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.

T = 200 N·md_o = 80 mmd_i = 60 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 75 kWN = 500 rpmτ_allow = 60 MPa

d = 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.

30 kN @ 10 m40 kN @ 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

Find the bending moment M_x at the section x = 8 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 10 kN at 8 mw = 3 kN/m from 0 to 10 mx = 8 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 100 mm wide and 16 mm thick, an overall depth of 240 mm and a web 6 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 100 mmH = 240 mmt_f = 16 mmt_w = 6 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 80 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 200 mmcut-out: 50 × 20 mmhole centre: y = 80 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 10 MPa, σᵧ = 50 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 10 MPaσ_y = 50 MPaτ_xy = −50 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −80 MPa, σᵧ = 10 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −80 MPaσ_y = 10 MPaτ_xy = −50 MPaθ = 45°

τ_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 = 6 kN/mL = 3 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 9 kNw = 12 kN/mL = 4 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 1 mE = 70 GPa (aluminium)b × h = 40 × 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 = 6 mE = 200 GPa (steel)d = 60 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 7.639 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·10²/4 = 78.54 mm²

σ = P/A = 40·1000/78.54 = 509.3 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 509.3/200000 = 0.002546 = 2546 µε

Elongation

δ = ε·L = 0.002546·3000 mm = 7.639 mm

Answer

δ = 7.639 mm

2.Axial stress & strain

Answer: P_max = 553.7 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 − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·5890 = 5.537 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001) = 0.001 (aluminium)

Largest load

P_max = 0.001·5.537 × 10⁸ = 553706 N = 553.7 kN

Answer

P_max = 553.7 kN

3.Torsion of shafts

Answer: φ = 0.4008 °

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(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Angle of twist

T = 200 N·m = 200000 N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 200000·2500/(26000·2.749 × 10⁶) = 0.006996 rad

φ = 0.006996·180/π = 0.4008°

Answer

φ = 0.4008°

4.Torsion of shafts

Answer: d = 49.54 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 75000/52.36 = 1432 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·1.432 × 10⁶/(π·60)) = 49.54 mm

Answer

d = 49.54 mm

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

5.Shear force & bending moment

Answer: E (x = 8 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 = 18.33 kN, R₂ = 51.67 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 = 36.67, M_C = 73.33, M_D = 110, M_E = 146.7 kN·m, M_F = 103.3, M_G = 0

Answer

Mark E: M_E = 146.7 kN·m

6.Shear force & bending moment

Answer: M_x = 40 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 = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 10·8 + 3·10·5

R_B = 230/10 = 23 kN

ΣF_y = 0: R_A = 40 − 23 = 17 kN

Cut at x = 8 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 = 17·8 − 3·8·4 = 40 kN·m

Answer

M_x = 40 kN·m

7.Centroid & second moment of area

Answer: I_x = 44.71 × 10⁶ mm⁴

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 = 240 − 2·16 = 208 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (100·240³ − 94·208³)/12

I_x = 4.471 × 10⁷ mm⁴ = 44.71 × 10⁶ mm⁴

Answer

I_x = 44.71 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 99.55 × 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 = (30000·100 − 1000·80)/(30000 − 1000) = 100.7 mm

Full plate about ȳ

150·200³/12 + 30000·−0.6897² = 100 × 10⁶ mm⁴

Hole about ȳ

50·20³/12 + 1000·−20.69² = 0.4614 × 10⁶ mm⁴

Subtract

I_x = 99.55 × 10⁶ mm⁴

Answer

I_x = 99.55 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 53.85 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 = √(((10 − 50)/2)² + −50²) = 53.85 MPa

Answer

τ_max = 53.85 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 45 MPa

Transformation equation

τ_θ = −−45·1 + −50·0 = 45 MPa

Answer

τ_θ = 45 MPa

11.Beam deflection

Answer: δ_max = 1.586 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 = 3000 mm

EI = 3.99 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·6·3000⁴/(384·3.99 × 10¹²) = 1.586 mm

Answer

δ_max = 1.586 mm

12.Beam deflection

Answer: δ_max = 4.643 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 = 9000 N, w = 12 N/mm, L = 4000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 9000·4000³/(48·1.12 × 10¹³) = 1.071 mm

δ_w = 5wL⁴/(384EI) = 5·12·4000⁴/(384·1.12 × 10¹³) = 3.571 mm

Add them

δ_max = 1.071 + 3.571 = 4.643 mm

Answer

δ_max = 4.643 mm

13.Column buckling

Answer: P_cr = 73.69 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 = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·426667/2000² = 73693 N = 73.69 kN

Answer

P_cr = 73.69 kN

14.Column buckling

Answer: σ_cr = 3.084 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 2·6000 = 12000 mm

KL/r = 12000/15 = 800

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa

Answer

σ_cr = 3.084 MPa

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