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 brass bar under the axial load.

P = 120 kNd = 20 mmL = 0.5 mE = 100 GPa

δ = mm

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 = 80 kNsegment 1: Ø30 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø20 mm, 0.5 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 80 mm, inner diameter 60 mm, length 0.8 m) is fixed at one end and twisted by a torque of 500 N·m at the other. Find the angle of twist φ in degrees.

T = 500 N·md_o = 80 mmd_i = 60 mmL = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 75 kW at 1000 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 75 kWN = 1000 rpmτ_allow = 40 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries two point loads. Find the maximum bending moment M_max in the beam.

L = 6 mP₁ = 10 kN at 4 mP₂ = 40 kN at 5 m

M_max = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 20 kN at 2 mw = 12 kN/m from 0 to 8 mx = 7 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 200 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 120 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 200 mmd = 30 mmhole centre: y = 120 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 240 mm plate has a rectangular cut-out 60 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 = 200 mmh = 240 mmcut-out: 60 × 50 mmhole centre: y = 156 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 160 MPa, σᵧ = −20 MPa and τₓᵧ = −10 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 160 MPaσ_y = −20 MPaτ_xy = −10 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −60 MPa, σᵧ = 50 MPa and τₓᵧ = −60 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = −60 MPaσ_y = 50 MPaτ_xy = −60 MPaσ_Y = 460 MPa

n =

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 = 30 kNa = 1 mL = 2 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2 m from support B. Find the maximum deflection δ_max of the beam.

P = 4 kNb = 2 mL = 8 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 1.91 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·20²/4 = 314.2 mm²

σ = P/A = 120·1000/314.2 = 382 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 382/100000 = 0.00382 = 3820 µε

Elongation

δ = ε·L = 0.00382·500 mm = 1.91 mm

Answer

δ = 1.91 mm

2.Axial stress & strain

Answer: σ_max = 254.6 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·20²/4 = 314.2 mm²

σ₁ = 113.2 MPa

σ₂ = 254.6 MPa

Largest stress

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

σ_max = 254.6 MPa

Answer

σ_max = 254.6 MPa

3.Torsion of shafts

Answer: φ = 0.3207 °

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(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Angle of twist

T = 500 N·m = 500000 N·mm

L = 0.8 m = 800 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 500000·800/(26000·2.749 × 10⁶) = 0.005597 rad

φ = 0.005597·180/π = 0.3207°

Answer

φ = 0.3207°

4.Torsion of shafts

Answer: d = 57.22 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 75000/104.7 = 716.2 N·m

Diameter for strength

T = 716.2 N·m = 716197 N·mm

d_τ = ∛(16·716197/(π·40)) = 45.01 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·716197/(π·39000·1.745 × 10⁻⁵))^1/4 = 57.22 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(45.01, 57.22) = 57.22 mm

Answer

d = 57.22 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_max = 40 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.

ΣM_A = 0: R_B·6 = 10·4 + 40·5

R_B = 240/6 = 40 kN

ΣF_y = 0: R_A = 50 − 40 = 10 kN

Moment under each load

Keep the part left of each load.

under P₁: M = R_A·4 = 10·4 = 40 kN·m

under P₂: M = R_B·1 = 40·1 = 40 kN·m

Largest value

M_max = 40 kN·m (under P₁)

Answer

M_max = 40 kN·m

6.Shear force & bending moment

Answer: M_x = 47 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 20·2 + 12·8·4

R_B = 424/8 = 53 kN

ΣF_y = 0: R_A = 116 − 53 = 63 kN

Cut at x = 7 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 = 63·7 − 20·(7 − 2) − 12·7·3.5 = 47 kN·m

Answer

M_x = 47 kN·m

7.Centroid & second moment of area

Answer: I_x = 133 × 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 = (40000·100 − 706.9·120)/(40000 − 706.9) = 99.64 mm

Full plate about ȳ

200·200³/12 + 40000·0.3598² = 133.3 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·20.36² = 0.3328 × 10⁶ mm⁴

Subtract

I_x = 133 × 10⁶ mm⁴

Answer

I_x = 133 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 225.6 × 10⁶ mm⁴

Centroid

ȳ = (48000·120 − 3000·156)/(48000 − 3000) = 117.6 mm

Full plate about ȳ

200·240³/12 + 48000·2.4² = 230.7 × 10⁶ mm⁴

Hole about ȳ

60·50³/12 + 3000·38.4² = 5.049 × 10⁶ mm⁴

Subtract

I_x = 225.6 × 10⁶ mm⁴

Answer

I_x = 225.6 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = −3.17 °

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−10, 160 − −20) = −6.34°

Angle on the element

θ_p = −3.17°

Answer

θ_p = −3.17°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 2.826

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = −5 MPa, R = 81.39 MPa

σ₁ = 76.39 MPa, σ₂ = −86.39 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 162.8 MPa

n = 460/162.8 = 2.826

Answer

n = 2.826

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_B = 10.42 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 = 30000 N, a = 1000 mm, L = 2000 mm

EI = 2.4 × 10¹² 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) = 4.167 mm

θ_P = Pa²/(2EI) = 0.00625 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 10.42 mm

Answer

δ_B = 10.42 mm

12.Beam deflection

Answer: δ_max = 5.07 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 = 4000 N, b = 2000 mm, L = 8000 mm

EI = 5.88 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((8000² − 2000²)/3) = 4472 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 5.07 mm

Answer

δ_max = 5.07 mm

13.Column buckling

Answer: P_cr = 11.07 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/2800² = 11074 N = 11.07 kN

Answer

P_cr = 11.07 kN

14.Column buckling

Answer: σ_cr = 49.35 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 0.5·5000 = 2500 mm

KL/r = 2500/12.5 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σ_cr = 49.35 MPa

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