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

P = 25 kNd = 16 mmL = 2 mE = 200 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 = 15 kNsegment 1: Ø35 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 75 mm and length 1 m; segment BC has diameter 30 mm and length 1.2 m. A torque of 200 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 200 N·md₁ = 75 mm, L₁ = 1 md₂ = 30 mm, L₂ = 1.2 m

τ_max = MPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 40 mm is fixed at A. A torque of 200 N·m acts at B, 1.2 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 3 m along. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT_B = 200 N·mT_C = 250 N·mAB = 1.2 mBC = 3 m

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.

30 kN @ 8 m25 kN @ 10 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 beam rests on a pin at A and a roller at B and overhangs B by 2.5 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 = 8.5 mw = 3 kN/m from 0 to 8.5 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 = 260 mmt_f = 10 mmt_w = 16 mmh_w = 260 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 300 mm plate has a rectangular cut-out 60 mm wide and 40 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: 60 × 40 mmhole centre: y = 180 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 120 MPa and τₓᵧ = 50 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 30 MPaσ_y = 120 MPaτ_xy = 50 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 8 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 500 mmt = 8 mmp = 3 MPaα = 55°

|τ_n| = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_B at the free end B.

P = 12 kNa = 1 mL = 6 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_B = 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 = 20 kN/mL = 2 mE = 11 GPa (timber)

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 = 1 mE = 70 GPa (aluminium)d = 20 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 621.7 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 = 25·1000/201.1 = 124.3 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 124.3/200000 = 0.0006217 = 621.7 µε

Answer

ε = 621.7 µε

2.Axial stress & strain

Answer: δ = 1.329 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·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 15.59 MPa

σ₂ = 132.6 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 15.59·1500/70000 = 0.3341 mm

δ₂ = σ₂L₂/E₂ = 132.6·1500/200000 = 0.9947 mm

δ = δ₁ + δ₂ = 1.329 mm

Answer

δ = 1.329 mm

3.Torsion of shafts

Answer: τ_max = 37.73 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 = 200 N·m = 200000 N·mm

τ_AB = 16·200000/(π·75³) = 2.414 MPa

τ_BC = 16·200000/(π·30³) = 37.73 MPa

Answer

τ_max = 37.73 MPa

4.Torsion of shafts

Answer: τ_max = 35.81 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 250 N·m

T_AB = T_C + T_B = 250 + 200 = 450 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 450 N·m

τ_max = 16·450000/(π·40³) = 35.81 MPa

Answer

τ_max = 35.81 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 = 14.17 kN, R₂ = 40.83 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 = 28.33, M_C = 56.67, M_D = 85, M_E = 113.3 kN·m, M_F = 81.67, M_G = 0

Answer

Mark E: M_E = 113.3 kN·m

6.Shear force & bending moment

Answer: M_max = 9.219 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 = 3·8.5 = 25.5 kN, acting at x̄ = 4.25 m

ΣM_A = 0: R_B·6 = 3·8.5·4.25

R_B = 108.4/6 = 18.06 kN

ΣF_y = 0: R_A = 25.5 − 18.06 = 7.438 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 7.438/3 = 2.479 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 7.438²/(2·3) = 9.219 kN·m

Answer

M_max = 9.219 kN·m

7.Centroid & second moment of area

Answer: I_x = 52.62 × 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 = 260·10 = 2600 mm², y_f = 260 + 10/2 = 265 mm

web: A_w = 16·260 = 4160 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2600·265 + 4160·130)/6760 = 181.9 mm

Parallel-axis theorem

flange: 260·10³/12 + 2600·(265 − 181.9)² = 1.797 × 10⁷ mm⁴

web: 16·260³/12 + 4160·(130 − 181.9)² = 3.465 × 10⁷ mm⁴

I_x = 1.797 × 10⁷ + 3.465 × 10⁷ = 5.262 × 10⁷ mm⁴ = 52.62 × 10⁶ mm⁴

Answer

I_x = 52.62 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 334.9 × 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 − 2400·180)/(45000 − 2400) = 148.3 mm

Full plate about ȳ

150·300³/12 + 45000·1.69² = 337.6 × 10⁶ mm⁴

Hole about ȳ

60·40³/12 + 2400·31.69² = 2.73 × 10⁶ mm⁴

Subtract

I_x = 334.9 × 10⁶ mm⁴

Answer

I_x = 334.9 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 71.13 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 75 MPa, R = √((−45)² + 50²) = 67.27 MPa

σ₁ = 142.3 MPa, σ₂ = 7.732 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(134.5, 142.3, 7.732)/2 = 71.13 MPa

Answer

τ_abs = 71.13 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 22.02 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. Thin-walled cylinder: hoop stress σ_h = pr/t and axial stress σ_a = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 500/2 = 250 mm

σ_h = pr/t = 3·250/8 = 93.75 MPa

σ_a = pr/(2t) = 46.88 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| = (93.75 − 46.88)·sin 55°·cos 55° = 22.02 MPa

Answer

|τ_n| = 22.02 MPa

11.Beam deflection

Answer: δ_B = 1.063 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 = 12000 N, a = 1000 mm, L = 6000 mm

EI = 3.2 × 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) = 0.125 mm

θ_P = Pa²/(2EI) = 0.0001875 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 1.063 mm

Answer

δ_B = 1.063 mm

12.Beam deflection

Answer: I_req = 68.18 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 2000/360 = 5.556 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·20·2000⁴/(384·11000·5.556) = 6.818 × 10⁷ mm⁴

I_req = 68.18 × 10⁶ mm⁴

Answer

I_req = 68.18 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 21.7 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 round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·1000 = 500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·7854/500² = 21704 N = 21.7 kN

Answer

P_cr = 21.7 kN

14.Column buckling

Answer: σ_cr = 22.56 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 0.7·5000 = 3500 mm

KL/r = 3500/20 = 175

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/175² = 22.56 MPa

Answer

σ_cr = 22.56 MPa

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