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 copper bar under the axial load.

P = 90 kNd = 16 mmL = 2.5 mE = 117 GPa

δ = mm

2.Axial stress & strainHard

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

P = 65 kNsegment 1: Ø25 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 2 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 25 mm and length 2 m; segment BC has diameter 20 mm and length 1 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 250 N·md₁ = 25 mm, L₁ = 2 md₂ = 20 mm, L₂ = 1 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 750 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 750 rpmd_o = 100 mmd_i = 70 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 @ 8 m25 kN @ 6 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

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment M_max in the beam.

L = 5 mw = 10 kN/m from 2 to 5 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 105 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 50 mmhole centre: y = 105 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 30 mm high, centred on the vertical centre line with its centre 105 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 150 mmcut-out: 80 × 30 mmhole centre: y = 105 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 30 MPa, σᵧ = −40 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 30 MPaσ_y = −40 MPaτ_xy = 60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 10 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 600 mmt = 10 mmp = 2.5 MPaα = 20°

|τ_n| = MPa

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 = 12 kN/mL = 1.5 mE = 70 GPa (aluminium)I = 8.5 × 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 = 2 kN/mL = 4 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)b × h = 20 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 85 kNn = 2.5L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 9.565 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·16²/4 = 201.1 mm²

σ = P/A = 90·1000/201.1 = 447.6 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 447.6/117000 = 0.003826 = 3826 µε

Elongation

δ = ε·L = 0.003826·2500 mm = 9.565 mm

Answer

δ = 9.565 mm

2.Axial stress & strain

Answer: δ = 3.895 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 132.4 MPa

σ₂ = 323.3 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 132.4·500/100000 = 0.6621 mm

δ₂ = σ₂L₂/E₂ = 323.3·2000/200000 = 3.233 mm

δ = δ₁ + δ₂ = 3.895 mm

Answer

δ = 3.895 mm

3.Torsion of shafts

Answer: τ_max = 159.2 MPa

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.

Stress in each segment

T = 250 N·m = 250000 N·mm

τ_AB = 16·250000/(π·25³) = 81.49 MPa

τ_BC = 16·250000/(π·20³) = 159.2 MPa

Answer

τ_max = 159.2 MPa

4.Torsion of shafts

Answer: τ_max = 6.4 MPa

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 75000/78.54 = 954.9 N·m

Polar second moment of area

J = π(100⁴ − 70⁴)/32 = 7.46 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·50/7.46 × 10⁶ = 6.4 MPa

Answer

τ_max = 6.4 MPa

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 = 32.5 kN, R₂ = 52.5 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 = 130, M_D = 195, M_E = 210 kN·m, M_F = 105, M_G = 0

Answer

Mark E: M_E = 210 kN·m

6.Shear force & bending moment

Answer: M_max = 22.05 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.

UDL resultant: W = w·ℓ = 10·3 = 30 kN, acting at x̄ = 3.5 m

ΣM_A = 0: R_B·5 = 10·3·3.5

R_B = 105/5 = 21 kN

ΣF_y = 0: R_A = 30 − 21 = 9 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 2) = 0

x₀ = 2 + 9/10 = 2.9 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 2)²/2 = 9·2.9 − 10·0.9²/2 = 22.05 kN·m

Answer

M_max = 22.05 kN·m

7.Centroid & second moment of area

Answer: I_x = 25.78 × 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 = (15000·75 − 1963·105)/(15000 − 1963) = 70.48 mm

Full plate about ȳ

100·150³/12 + 15000·4.518² = 28.43 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·34.52² = 2.646 × 10⁶ mm⁴

Subtract

I_x = 25.78 × 10⁶ mm⁴

Answer

I_x = 25.78 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 39.59 × 10⁶ mm⁴

Centroid

ȳ = (22500·75 − 2400·105)/(22500 − 2400) = 71.42 mm

Full plate about ȳ

150·150³/12 + 22500·3.582² = 42.48 × 10⁶ mm⁴

Hole about ȳ

80·30³/12 + 2400·33.58² = 2.887 × 10⁶ mm⁴

Subtract

I_x = 39.59 × 10⁶ mm⁴

Answer

I_x = 39.59 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 69.46 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 = √(((30 − −40)/2)² + 60²) = 69.46 MPa

Answer

τ_max = 69.46 MPa

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 12.05 MPa

Wall stresses

r = 600/2 = 300 mm

σ_h = pr/t = 2.5·300/10 = 75 MPa

σ_a = pr/(2t) = 37.5 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.

|τ_n| = (75 − 37.5)·sin 20°·cos 20° = 12.05 MPa

Answer

|τ_n| = 12.05 MPa

11.Beam deflection

Answer: θ_A = 0.002836 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 = 12 N/mm, L = 1500 mm

EI = 5.95 × 10¹¹ N·mm²

End slope

θ_A = wL³/(24EI) = 12·1500³/(24·5.95 × 10¹¹) = 0.002836 rad

Answer

θ_A = 0.002836 rad

That is 0.1625°.

12.Beam deflection

Answer: δ_max = 25.25 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 = 2 N/mm, L = 4000 mm, EI = 1.32 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 20000·4000³/(48·1.32 × 10¹²) = 20.2 mm

δ_w = 5wL⁴/(384EI) = 5·2·4000⁴/(384·1.32 × 10¹²) = 5.051 mm

Add them

δ_max = 20.2 + 5.051 = 25.25 mm

Answer

δ_max = 25.25 mm

13.Column buckling

Answer: P_cr = 11.7 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·53333/3000² = 11697 N = 11.7 kN

Answer

P_cr = 11.7 kN

14.Column buckling

Answer: d = 93.6 mm

Required critical load

P_cr = n·P = 2.5·85000 = 212500 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·3500²/(π³·70000))^(1/4) = 93.6 mm

Answer

d = 93.6 mm

Check: the slenderness KL/r is 149.6, so the column is slender and Euler applies.

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