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 = 75 kNd = 30 mmL = 1.5 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 = 75 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 0.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 75 N·m. The two ends of a gauge length of 300 mm rotate 7.92° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 75 N·md = 12 mmL = 300 mmφ = 7.92°

G = GPa

4.Torsion of shaftsHard

A hollow shaft must transmit 75 kW at 1000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter d_o.

P = 75 kWN = 1000 rpmd_i/d_o = 0.6τ_allow = 80 MPa

d_o = mm

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.

40 kN @ 2 m12 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

The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x₀ from A to the section where the bending moment is largest.

L = 4 mw = 6 kN/m from 1 to 3 m

x₀ = 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 = 280 mmt_f = 20 mmt_w = 10 mmh_w = 300 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 30 mm wide and 20 mm high, centred on the vertical centre line with its centre 144 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmcut-out: 30 × 20 mmhole centre: y = 144 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 40 MPa, σᵧ = 50 MPa and τₓᵧ = −30 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 40 MPaσ_y = 50 MPaτ_xy = −30 MPaθ = 35°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 100 N·m together with a torque of 250 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 40 mmM = 100 N·mT = 250 N·m

σ₁ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 2 m from support A. Find the deflection δ_P under the load.

P = 22 kNa = 2 mL = 5 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 4 mw = 8 kN/m

M_A = kN·m

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 = 2.5 mE = 70 GPa (aluminium)d = 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 65 kNn = 2.5L = 6 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 906.9 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·30²/4 = 706.9 mm²

σ = P/A = 75·1000/706.9 = 106.1 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 106.1/117000 = 0.0009069 = 906.9 µε

Answer

ε = 906.9 µε

2.Axial stress & strain

Answer: δ = 2.489 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·40²/4 = 1257 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 59.68 MPa

σ₂ = 373 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 59.68·1500/100000 = 0.8952 mm

δ₂ = σ₂L₂/E₂ = 373·500/117000 = 1.594 mm

δ = δ₁ + δ₂ = 2.489 mm

Answer

δ = 2.489 mm

3.Torsion of shafts

Answer: G = 79.96 GPa

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 moment and angle

J = π·12⁴/32 = 2036 mm⁴

φ = 7.92·π/180 = 0.1382 rad

Shear modulus

T = 75 N·m = 75000 N·mm

G = TL/(Jφ) = 75000·300/(2036·0.1382) = 79957 MPa

G = 79.96 GPa

Answer

G = 79.96 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d_o = 37.42 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 75000/104.7 = 716.2 N·m

Outer diameter from the allowable stress

T = 716.2 N·m = 716197 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·716197/(π·80·0.8704)) = 37.42 mm

d_i = 0.6·37.42 = 22.45 mm

Answer

d_o = 37.42 mm

A solid shaft would need d = 35.72 mm; the hollow one is a little larger but much lighter.

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 = 62.67 kN, R₂ = 49.33 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 = 56.67, M_C = 101.3 kN·m, M_D = 94, M_E = 74.67, M_F = 43.33, M_G = 0

Answer

Mark C: M_C = 101.3 kN·m

6.Shear force & bending moment

Answer: x₀ = 2 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 = 6·2 = 12 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 6·2·2

R_B = 24/4 = 6 kN

ΣF_y = 0: R_A = 12 − 6 = 6 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 − 1) = 0

x₀ = 1 + 6/6 = 2 m

Answer

x₀ = 2 m

7.Centroid & second moment of area

Answer: I_x = 72.7 × 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 = 280·20 = 5600 mm², y_f = 300 + 20/2 = 310 mm

web: A_w = 10·300 = 3000 mm², y_w = 300/2 = 150 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (5600·310 + 3000·150)/8600 = 254.2 mm

Parallel-axis theorem

flange: 280·20³/12 + 5600·(310 − 254.2)² = 1.763 × 10⁷ mm⁴

web: 10·300³/12 + 3000·(150 − 254.2)² = 5.506 × 10⁷ mm⁴

I_x = 1.763 × 10⁷ + 5.506 × 10⁷ = 7.27 × 10⁷ mm⁴ = 72.7 × 10⁶ mm⁴

Answer

I_x = 72.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 137.9 × 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 = (28800·120 − 600·144)/(28800 − 600) = 119.5 mm

Full plate about ȳ

120·240³/12 + 28800·0.5106² = 138.2 × 10⁶ mm⁴

Hole about ȳ

30·20³/12 + 600·24.51² = 0.3805 × 10⁶ mm⁴

Subtract

I_x = 137.9 × 10⁶ mm⁴

Answer

I_x = 137.9 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 74.9 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.

Transformation equation for the y′ face

σ_y′ = 45 − (−5)·0.342 − (−30)·0.9397 = 74.9 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 15.1 MPa, σ_x′ + σ_y′ = 90 = σ_x + σ_y

Answer

σ_y′ = 74.9 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 29.38 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·100000/(π·40³) = 15.92 MPa

τ = 16T/(πd³) = 16·250000/(π·40³) = 19.89 MPa

Mohr’s circle

C = σ/2 = 7.958 MPa

R = √(7.958² + 19.89²) = 21.43 MPa

σ₁ = C + R = 29.38 MPa

σ₂ = C − R = −13.47 MPa

Answer

σ₁ = 29.38 MPa

11.Beam deflection

Answer: δ_P = 13.23 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 = 22000 N, a = 2000 mm, b = 3000 mm, L = 5000 mm

EI = 3.99 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 22000·2000²·3000²/(3·3.99 × 10¹²·5000) = 13.23 mm

Answer

δ_P = 13.23 mm

12.Beam deflection

Answer: M_A = 16 kN·m

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·8·4/8 = 12 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = wL²/2 − R_B·L = 64 − 48 = 16 kN·m

Answer

M_A = 16 kN·m

13.Column buckling

Answer: P_cr = 135.7 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/1250² = 135652 N = 135.7 kN

Answer

P_cr = 135.7 kN

14.Column buckling

Answer: d = 81.04 mm

Required critical load

P_cr = n·P = 2.5·65000 = 162500 N

KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·3000²/(π³·70000))^(1/4) = 81.04 mm

Answer

d = 81.04 mm

Check: the slenderness KL/r is 148.1, so the column is slender and Euler applies.

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