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

P = 75 kNd = 16 mmL = 0.5 mE = 70 GPa

ε = µε

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 = 20 kNsegment 1: Ø25 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 100 N·md = 30 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 5 kW at 1800 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

P = 5 kWN = 1800 rpmd_i/d_o = 0.75τ_allow = 50 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.

25 kN @ 4 m45 kN @ 2 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 = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 15 kN at 2.5 mw = 6 kN/m from 0 to 6 mx = 2.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 240 mm plate has a circular hole of diameter 30 mm on its 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 mmd = 30 mmhole centre: y = 72 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 90 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 300 mmcut-out: 50 × 20 mmhole centre: y = 90 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −80 MPa, σᵧ = 0 MPa and τₓᵧ = 50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −80 MPaσ_y = 0 MPaτ_xy = 50 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 0 MPa, σᵧ = −80 MPa and τₓᵧ = 30 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 0 MPaσ_y = −80 MPaτ_xy = 30 MPa

σ_vM = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2 m from A. Find the deflection δ_B at the free end B.

P = 16 kNa = 2 mL = 2.5 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 8 mP = 50 kN at 4 m

R_B = kN

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 = 30 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 60 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 60 kNn = 2.5L = 6 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 5329 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·16²/4 = 201.1 mm²

σ = P/A = 75·1000/201.1 = 373 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 373/70000 = 0.005329 = 5329 µε

Answer

ε = 5329 µε

2.Axial stress & strain

Answer: σ_max = 176.8 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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 40.74 MPa

σ₂ = 176.8 MPa

Largest stress

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

σ_max = 176.8 MPa

Answer

σ_max = 176.8 MPa

3.Torsion of shafts

Answer: φ = 2.771 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1500/(39000·79522) = 0.04837 rad

φ = 0.04837·180/π = 2.771°

Answer

φ = 2.771°

4.Torsion of shafts

Answer: d_o = 15.81 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 5000/188.5 = 26.53 N·m

Outer diameter from the allowable stress

T = 26.53 N·m = 26526 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·26526/(π·50·0.6836)) = 15.81 mm

d_i = 0.75·15.81 = 11.86 mm

Answer

d_o = 15.81 mm

A solid shaft would need d = 13.93 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: C (x = 2 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 = 38.33 kN, R₂ = 31.67 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 = 38.33, M_C = 76.67 kN·m, M_D = 70, M_E = 63.33, M_F = 31.67, M_G = 0

Answer

Mark C: M_C = 76.67 kN·m

6.Shear force & bending moment

Answer: M_x = 48.13 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 = 6·6 = 36 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 15·2.5 + 6·6·3

R_B = 145.5/6 = 24.25 kN

ΣF_y = 0: R_A = 51 − 24.25 = 26.75 kN

Cut at x = 2.5 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 = 26.75·2.5 − 6·2.5·1.25 = 48.13 kN·m

Answer

M_x = 48.13 kN·m

7.Centroid & second moment of area

Answer: I_x = 136.5 × 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 − 706.9·72)/(28800 − 706.9) = 121.2 mm

Full plate about ȳ

120·240³/12 + 28800·−1.208² = 138.3 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−49.21² = 1.751 × 10⁶ mm⁴

Subtract

I_x = 136.5 × 10⁶ mm⁴

Answer

I_x = 136.5 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 266.3 × 10⁶ mm⁴

Centroid

ȳ = (36000·150 − 1000·90)/(36000 − 1000) = 151.7 mm

Full plate about ȳ

120·300³/12 + 36000·−1.714² = 270.1 × 10⁶ mm⁴

Hole about ȳ

50·20³/12 + 1000·−61.71² = 3.842 × 10⁶ mm⁴

Subtract

I_x = 266.3 × 10⁶ mm⁴

Answer

I_x = 266.3 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 64.03 MPa

Radius of Mohr’s circle

τ_max = R = √(((−80 − 0)/2)² + 50²) = 64.03 MPa

Answer

τ_max = 64.03 MPa

10.Mohr’s circle & stress transformation

Answer: σ_vM = 95.39 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 = −40 MPa, R = 50 MPa

σ₁ = 10 MPa, σ₂ = −90 MPa, σ₃ = 0

Von Mises

σ_vM = √(10² − 10·(−90) + (−90)²) = 95.39 MPa

Answer

σ_vM = 95.39 MPa

11.Beam deflection

Answer: δ_B = 24.44 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 = 16000 N, a = 2000 mm, L = 2500 mm

EI = 2.4 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 17.78 mm

θ_P = Pa²/(2EI) = 0.01333 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 24.44 mm

Answer

δ_B = 24.44 mm

12.Beam deflection

Answer: R_B = 15.63 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = 5PL³/(48EI) → R_B = 5P/16 = 5·50/16 = 15.63 kN

Answer

R_B = 15.63 kN

13.Column buckling

Answer: P_cr = 23.32 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·135000/2000² = 23317 N = 23.32 kN

Answer

P_cr = 23.32 kN

14.Column buckling

Answer: d = 61.1 mm

Required critical load

P_cr = n·P = 2.5·60000 = 150000 N

KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·3000²/(π³·200000))^(1/4) = 61.1 mm

Answer

d = 61.1 mm

Check: the slenderness KL/r is 196.4, so the column is slender and Euler applies.

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