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 = 95 kNd = 12 mmL = 2 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 120 kNsegment 1: Ø30 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel 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 = 250 N·md = 40 mmL = 0.8 mG = 80 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 80 MPa. Find the smallest outer diameter d_o.

P = 5 kWN = 1800 rpmd_i/d_o = 0.75τ_allow = 80 MPa

d_o = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 55 kN·m at x = 2 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 55 kN·m clockwise, at 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 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 7.5 mw = 12 kN/m from 0 to 7.5 msupports at x = 0 and x = 6 m

M_max = kN·m

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 300 mmt_f = 10 mmt_w = 8 mmh_w = 160 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 300 mm plate has a rectangular cut-out 40 mm wide and 50 mm high, centred on the vertical centre line with its centre 180 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 300 mmcut-out: 40 × 50 mmhole centre: y = 180 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa and σᵧ = 90 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 135 MPa?

σ_x = −30 MPaσ_y = 90 MPaσ₁ = 135 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = −40 MPa and τₓᵧ = 60 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 30 MPaσ_y = −40 MPaτ_xy = 60 MPaσ_Y = 350 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

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

θ_A = rad

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 = 20 kNw = 6 kN/mL = 5 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 2 mE = 70 GPa (aluminium)b × h = 40 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 100 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 100 kNn = 3L = 2.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 7179 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·12²/4 = 113.1 mm²

σ = P/A = 95·1000/113.1 = 840 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 840/117000 = 0.007179 = 7179 µε

Answer

ε = 7179 µε

2.Axial stress & strain

Answer: δ = 10.63 mm

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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 169.8 MPa

σ₂ = 596.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 169.8·500/200000 = 0.4244 mm

δ₂ = σ₂L₂/E₂ = 596.8·2000/117000 = 10.2 mm

δ = δ₁ + δ₂ = 10.63 mm

Answer

δ = 10.63 mm

3.Torsion of shafts

Answer: φ = 0.5699 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·800/(80000·251327) = 0.009947 rad

φ = 0.009947·180/π = 0.5699°

Answer

φ = 0.5699°

4.Torsion of shafts

Answer: d_o = 13.52 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·80·0.6836)) = 13.52 mm

d_i = 0.75·13.52 = 10.14 mm

Answer

d_o = 13.52 mm

A solid shaft would need d = 11.91 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_x = −18.33 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 55/6 = 9.167 kN (upward)

R_A = −R_B = −9.167 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −9.167·2 = −18.33 kN·m

Answer

M_x = −18.33 kN·m

6.Shear force & bending moment

Answer: M_max = 47.46 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 12·7.5 = 90 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·6 = 12·7.5·3.75

R_B = 337.5/6 = 56.25 kN

ΣF_y = 0: R_A = 90 − 56.25 = 33.75 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 33.75/12 = 2.813 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 33.75²/(2·12) = 47.46 kN·m

Answer

M_max = 47.46 kN·m

7.Centroid & second moment of area

Answer: I_x = 9.238 × 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 two rectangles

Measure every y from the bottom edge.

flange: A_f = 300·10 = 3000 mm², y_f = 160 + 10/2 = 165 mm

web: A_w = 8·160 = 1280 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·165 + 1280·80)/4280 = 139.6 mm

Parallel-axis theorem

flange: 300·10³/12 + 3000·(165 − 139.6)² = 1.964 × 10⁶ mm⁴

web: 8·160³/12 + 1280·(80 − 139.6)² = 7.274 × 10⁶ mm⁴

I_x = 1.964 × 10⁶ + 7.274 × 10⁶ = 9.238 × 10⁶ mm⁴ = 9.238 × 10⁶ mm⁴

Answer

I_x = 9.238 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 335.2 × 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 = (45000·150 − 2000·180)/(45000 − 2000) = 148.6 mm

Full plate about ȳ

150·300³/12 + 45000·1.395² = 337.6 × 10⁶ mm⁴

Hole about ȳ

40·50³/12 + 2000·31.4² = 2.388 × 10⁶ mm⁴

Subtract

I_x = 335.2 × 10⁶ mm⁴

Answer

I_x = 335.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 86.17 MPa

Required radius

C = (−30 + 90)/2 = 30 MPa

R = σ₁ − C = 135 − 30 = 105 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(105² − 60²) = 86.17 MPa

Answer

τ_xy = 86.17 MPa

10.Mohr’s circle & stress transformation

Answer: n = 2.907

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 = −5 MPa, R = 69.46 MPa

σ₁ = 64.46 MPa, σ₂ = −74.46 MPa, σ₃ = 0

Von Mises

σ_vM = √(64.46² − 64.46·(−74.46) + (−74.46)²) = 120.4 MPa

n = 350/120.4 = 2.907

Answer

n = 2.907

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: θ_A = 0.01353 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 6000 mm

EI = 3.99 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 6·6000³/(24·3.99 × 10¹²) = 0.01353 rad

Answer

θ_A = 0.01353 rad

That is 0.7754°.

12.Beam deflection

Answer: δ_max = 76.45 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 = 20000 N, w = 6 N/mm, L = 5000 mm, EI = 1.32 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 20000·5000³/(48·1.32 × 10¹²) = 39.46 mm

δ_w = 5wL⁴/(384EI) = 5·6·5000⁴/(384·1.32 × 10¹²) = 36.99 mm

Add them

δ_max = 39.46 + 36.99 = 76.45 mm

Answer

δ_max = 76.45 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 = 1·2000 = 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: d = 93.8 mm

Required critical load

P_cr = n·P = 3·100000 = 300000 N

KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·5000²/(π³·200000))^(1/4) = 93.8 mm

Answer

d = 93.8 mm

Check: the slenderness KL/r is 213.2, so the column is slender and Euler applies.

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