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

P = 85 kNd = 10 mmL = 1 mE = 100 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 = 35 kNsegment 1: Ø30 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 75 mm and length 3 m; segment BC has diameter 50 mm and length 3 m. A torque of 200 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 200 N·md₁ = 75 mm, L₁ = 3 md₂ = 50 mm, L₂ = 3 mG = 39 GPa

φ_C = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1450 rpmd_o = 50 mmd_i = 35 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.

60 kN @ 2 m10 kN/m, 0–6 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 = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 40 kN at 2.5 mw = 12 kN/m from 0 to 4 mx = 1 m

M_x = 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 = 120 mmt_f = 15 mmt_w = 16 mmh_w = 100 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 100 mm × 240 mm plate has a rectangular cut-out 30 mm wide and 50 mm high, centred on the vertical centre line with its centre 156 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 240 mmcut-out: 30 × 50 mmhole centre: y = 156 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 60 MPa and σᵧ = 20 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 105 MPa?

σ_x = 60 MPaσ_y = 20 MPaσ₁ = 105 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 15 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 15 mmp = 3 MPaα = 50°

σ_n = 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 = 2 kN/mL = 5 mE = 70 GPa (aluminium)I = 120 × 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 = 4 kN/mL = 4 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

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)b × h = 25 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 125 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 125 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 10823 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·10²/4 = 78.54 mm²

σ = P/A = 85·1000/78.54 = 1082 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 1082/100000 = 0.01082 = 10823 µε

Answer

ε = 10823 µε

2.Axial stress & strain

Answer: δ = 5.382 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·12²/4 = 113.1 mm²

σ₁ = 49.51 MPa

σ₂ = 309.5 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 49.51·2000/70000 = 1.415 mm

δ₂ = σ₂L₂/E₂ = 309.5·1500/117000 = 3.968 mm

δ = δ₁ + δ₂ = 5.382 mm

Answer

δ = 5.382 mm

3.Torsion of shafts

Answer: φ_C = 1.72 °

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 moments

J₁ = π·75⁴/32 = 3.106 × 10⁶ mm⁴

J₂ = π·50⁴/32 = 613592 mm⁴

Twist of each segment

T = 200 N·m = 200000 N·mm

G = 39 GPa = 39000 MPa

φ_AB = 200000·3000/(39000·3.106 × 10⁶) = 0.004953 rad

φ_BC = 200000·3000/(39000·613592) = 0.02507 rad

Total twist

φ_C = 0.004953 + 0.02507 = 0.03003 rad = 1.72°

Answer

φ_C = 1.72°

4.Torsion of shafts

Answer: τ_max = 35.31 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(50⁴ − 35⁴)/32 = 466269 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 658572·25/466269 = 35.31 MPa

Answer

τ_max = 35.31 MPa

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 = 70 kN, R₂ = 50 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 = 65, M_C = 120 kN·m, M_D = 105, M_E = 80, M_F = 45, M_G = 0

Answer

Mark C: M_C = 120 kN·m

6.Shear force & bending moment

Answer: M_x = 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·4 = 48 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 40·2.5 + 12·4·2

R_B = 196/4 = 49 kN

ΣF_y = 0: R_A = 88 − 49 = 39 kN

Cut at x = 1 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 = 39·1 − 12·1·0.5 = 33 kN·m

Answer

M_x = 33 kN·m

7.Centroid & second moment of area

Answer: I_x = 4.168 × 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 = 120·15 = 1800 mm², y_f = 100 + 15/2 = 107.5 mm

web: A_w = 16·100 = 1600 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1800·107.5 + 1600·50)/3400 = 80.44 mm

Parallel-axis theorem

flange: 120·15³/12 + 1800·(107.5 − 80.44)² = 1.352 × 10⁶ mm⁴

web: 16·100³/12 + 1600·(50 − 80.44)² = 2.816 × 10⁶ mm⁴

I_x = 1.352 × 10⁶ + 2.816 × 10⁶ = 4.168 × 10⁶ mm⁴ = 4.168 × 10⁶ mm⁴

Answer

I_x = 4.168 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 112.8 × 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 = (24000·120 − 1500·156)/(24000 − 1500) = 117.6 mm

Full plate about ȳ

100·240³/12 + 24000·2.4² = 115.3 × 10⁶ mm⁴

Hole about ȳ

30·50³/12 + 1500·38.4² = 2.524 × 10⁶ mm⁴

Subtract

I_x = 112.8 × 10⁶ mm⁴

Answer

I_x = 112.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 61.85 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.

Required radius

C = (60 + 20)/2 = 40 MPa

R = σ₁ − C = 105 − 40 = 65 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 20²) = 61.85 MPa

Answer

τ_xy = 61.85 MPa

10.Mohr’s circle & stress transformation

Answer: σ_n = 70.66 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 3·500/15 = 100 MPa

σ_a = pr/(2t) = 50 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 50·sin²50° + 100·cos²50° = 70.66 MPa

Answer

σ_n = 70.66 MPa

11.Beam deflection

Answer: δ_max = 1.938 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 = 2 N/mm, L = 5000 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·2·5000⁴/(384·8.4 × 10¹²) = 1.938 mm

Answer

δ_max = 1.938 mm

12.Beam deflection

Answer: I_req = 17.14 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·4·4000⁴/(384·70000·11.11) = 1.714 × 10⁷ mm⁴

I_req = 17.14 × 10⁶ mm⁴

Answer

I_req = 17.14 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 11.51 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·104167/2500² = 11515 N = 11.51 kN

Answer

P_cr = 11.51 kN

14.Column buckling

Answer: d = 84.76 mm

Required critical load

P_cr = n·P = 2.5·125000 = 312500 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·312500·4000²/(π³·200000))^(1/4) = 84.76 mm

Answer

d = 84.76 mm

Check: the slenderness KL/r is 188.8, so the column is slender and Euler applies.

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