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

A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaδ = 0.5 mm

P = kN

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 = 15 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 400 N·md = 50 mmL = 0.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 30 kW at 1450 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

P = 30 kWN = 1450 rpmd_i/d_o = 0.8τ_allow = 60 MPa

d_o = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 8 kN/m from 0 to 5 mx = 3 m

V_x = kN

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 25 kN at 3.5 mw = 5 kN/m from 0 to 5 mx = 3.5 m

M_x = 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, 160 mm high overall. Base and legs are all 10 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 160 mmh = 160 mmt = 10 mm

ȳ = mm

8.Centroid & second moment of areaHard

A 200 mm × 150 mm plate has a rectangular cut-out 30 mm wide and 30 mm high, centred on the vertical centre line with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 150 mmcut-out: 30 × 30 mmhole centre: y = 45 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −50 MPa, σᵧ = 10 MPa and τₓᵧ = 60 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = −50 MPaσ_y = 10 MPaτ_xy = 60 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σ_x = 160 MPaσ_y = −60 MPaτ_xy = 30 MPaθ = 25°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 3 m from A. Find the deflection δ_B at the free end B.

P = 14 kNa = 3 mL = 5 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δ_max.

P = 15 kNL = 4 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is pinned at both ends. It must carry 35 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 35 kNn = 2.5L = 6 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 78.54 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

A₁ = 706.9 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(100000·706.9) = 2.829 × 10⁻⁶ mm/N

Load

P = 0.5/6.366 × 10⁻⁶ = 78540 N = 78.54 kN

Answer

P = 78.54 kN

2.Axial stress & strain

Answer: σ_max = 132.6 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·30²/4 = 706.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 21.22 MPa

σ₂ = 132.6 MPa

Largest stress

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

σ_max = 132.6 MPa

Answer

σ_max = 132.6 MPa

3.Torsion of shafts

Answer: φ = 0.7183 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·500/(26000·613592) = 0.01254 rad

φ = 0.01254·180/π = 0.7183°

Answer

φ = 0.7183°

4.Torsion of shafts

Answer: d_o = 30.51 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 30000/151.8 = 197.6 N·m

Outer diameter from the allowable stress

T = 197.6 N·m = 197572 N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·197572/(π·60·0.5904)) = 30.51 mm

d_i = 0.8·30.51 = 24.41 mm

Answer

d_o = 30.51 mm

A solid shaft would need d = 25.6 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: V_x = −4 kN

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 = 8·5 = 40 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 8·5·2.5

R_B = 100/5 = 20 kN

ΣF_y = 0: R_A = 40 − 20 = 20 kN

Cut at x = 3 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.

V = R_A − (all downward loads left of the cut)

V_x = 20 − 8·3 = −4 kN

Answer

V_x = −4 kN

6.Shear force & bending moment

Answer: M_x = 39.38 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·5 = 25 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 25·3.5 + 5·5·2.5

R_B = 150/5 = 30 kN

ΣF_y = 0: R_A = 50 − 30 = 20 kN

Cut at x = 3.5 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 = 20·3.5 − 5·3.5·1.75 = 39.38 kN·m

Answer

M_x = 39.38 kN·m

7.Centroid & second moment of area

Answer: ȳ = 57.17 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 = 160·10 = 1600 mm², y₁ = 5 mm

two legs: A₂ = 2·10·150 = 3000 mm², y₂ = 10 + 150/2 = 85 mm

Centroid

ȳ = (1600·5 + 3000·85)/4600 = 57.17 mm

Answer

ȳ = 57.17 mm

8.Centroid & second moment of area

Answer: I_x = 55.35 × 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 = (30000·75 − 900·45)/(30000 − 900) = 75.93 mm

Full plate about ȳ

200·150³/12 + 30000·−0.9278² = 56.28 × 10⁶ mm⁴

Hole about ȳ

30·30³/12 + 900·−30.93² = 0.9284 × 10⁶ mm⁴

Subtract

I_x = 55.35 × 10⁶ mm⁴

Answer

I_x = 55.35 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = 13.28 °

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.

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 60/67.08 = 0.8944, sin 2θ_s = −(−30)/67.08 = 0.4472

2θ_s = atan2(30, 60) = 26.57°

Angle on the element

θ_s = 13.28°

check: θ_s − θ_p = 13.28 − 58.28 = −45°

Answer

θ_s = 13.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 143.7 MPa

Transformation equation

σ_θ = 50 + 110·0.6428 + 30·0.766 = 143.7 MPa

Answer

σ_θ = 143.7 MPa

11.Beam deflection

Answer: δ_B = 15 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 = 14000 N, a = 3000 mm, L = 5000 mm

EI = 1.68 × 10¹³ 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) = 7.5 mm

θ_P = Pa²/(2EI) = 0.00375 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 15 mm

Answer

δ_B = 15 mm

12.Beam deflection

Answer: δ_max = 1.065 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 = 15000 N, L = 4000 mm

EI = 3.2 × 10¹³ N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 1333 mm

δ₁ = 0.5324 mm

Add both loads

δ_max = 2·δ₁ = 23PL³/(648EI) = 1.065 mm

Answer

δ_max = 1.065 mm

13.Column buckling

Answer: P_cr = 70.32 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 round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·636173/2500² = 70322 N = 70.32 kN

Answer

P_cr = 70.32 kN

14.Column buckling

Answer: d = 98.17 mm

Required critical load

P_cr = n·P = 2.5·35000 = 87500 N

KL = 1·6000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·87500·6000²/(π³·70000))^(1/4) = 98.17 mm

Answer

d = 98.17 mm

Check: the slenderness KL/r is 244.5, so the column is slender and Euler applies.

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