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 = 45 kNd = 20 mmL = 1.5 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 total elongation δ.

P = 10 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, brass (E = 100 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 50 mm, inner diameter 25 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 = 50 mmd_i = 25 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 750 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 20 kWN = 750 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 35 kN·m at x = 1 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 5 mM₀ = 35 kN·m clockwise, at 1 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment M_B at support B. Use sagging moments positive and upward shear on the left face positive.

L = 9.5 mw = 6 kN/m from 0 to 9.5 msupports at x = 0 and x = 8 m

M_B = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 30 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 150 mm wide and 24 mm thick, an overall depth of 200 mm and a web 14 mm thick. It carries a bending moment of 120 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 150 mmH = 200 mmt_f = 24 mmt_w = 14 mmM = 120 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 50 MPa and σᵧ = 30 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 65 MPa?

σ_x = 50 MPaσ_y = 30 MPaσ₁ = 65 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 0 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.

σ_x = 0 MPaσ_y = −60 MPaτ_xy = 10 MPaθ = 45°

σ_θ = 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 = 10 kN/mL = 2 mE = 70 GPa (aluminium)I = 12 × 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 = 6 kN/mL = 6 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

14.Column bucklingHard

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

L = 5 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 2.149 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 = 45·1000/314.2 = 143.2 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 143.2/100000 = 0.001432 = 1432 µε

Elongation

δ = ε·L = 0.001432·1500 mm = 2.149 mm

Answer

δ = 2.149 mm

2.Axial stress & strain

Answer: δ = 0.3926 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·20²/4 = 314.2 mm²

σ₁ = 10.39 MPa

σ₂ = 31.83 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 10.39·500/70000 = 0.07424 mm

δ₂ = σ₂L₂/E₂ = 31.83·1000/100000 = 0.3183 mm

δ = δ₁ + δ₂ = 0.3926 mm

Answer

δ = 0.3926 mm

3.Torsion of shafts

Answer: φ = 1.022 °

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(50⁴ − 25⁴)/32 = 575243 mm⁴

Angle of twist

T = 500 N·m = 500000 N·mm

L = 0.8 m = 800 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 500000·800/(39000·575243) = 0.01783 rad

φ = 0.01783·180/π = 1.022°

Answer

φ = 1.022°

4.Torsion of shafts

Answer: d = 29.6 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 20000/78.54 = 254.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·254648/(π·50)) = 29.6 mm

Answer

d = 29.6 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_x = −7 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 35/5 = 7 kN (upward)

R_A = −R_B = −7 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −7·1 = −7 kN·m

Answer

M_x = −7 kN·m

6.Shear force & bending moment

Answer: M_B = −6.75 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 = 6·9.5 = 57 kN, acting at x̄ = 4.75 m

ΣM_A = 0: R_B·8 = 6·9.5·4.75

R_B = 270.8/8 = 33.84 kN

ΣF_y = 0: R_A = 57 − 33.84 = 23.16 kN

Cut at B and keep the overhang

The load on the overhang, w·a, acts at a/2 from B and makes the beam hog over the support.

M_B = −w·a²/2 = −6·1.5²/2 = −6.75 kN·m

Answer

M_B = −6.75 kN·m

7.Centroid & second moment of area

Answer: I_x = 27.71 × 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 − 706.9·52.5)/(15000 − 706.9) = 76.11 mm

Full plate about ȳ

100·150³/12 + 15000·−1.113² = 28.14 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−23.61² = 0.4339 × 10⁶ mm⁴

Subtract

I_x = 27.71 × 10⁶ mm⁴

Answer

I_x = 27.71 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_j = 151.5 MPa

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.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 200 − 2·24 = 152 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (150·200³ − 136·152³)/12

I_x = 6.02 × 10⁷ mm⁴

Bending stress

y = 100 − 24 = 76 mm

σ = M·y/I = 1.2 × 10⁸·76/6.02 × 10⁷ = 151.5 MPa

Answer

σ_j = 151.5 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 22.91 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 = (50 + 30)/2 = 40 MPa

R = σ₁ − C = 65 − 40 = 25 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(25² − 10²) = 22.91 MPa

Answer

τ_xy = 22.91 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −20 MPa

Transformation equation

σ_θ = −30 + 30·0 + 10·1 = −20 MPa

Answer

σ_θ = −20 MPa

11.Beam deflection

Answer: δ_max = 2.48 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 = 10 N/mm, L = 2000 mm

EI = 8.4 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·2000⁴/(384·8.4 × 10¹¹) = 2.48 mm

Answer

δ_max = 2.48 mm

12.Beam deflection

Answer: I_req = 86.79 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 6000/360 = 16.67 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·6·6000⁴/(384·70000·16.67) = 8.679 × 10⁷ mm⁴

I_req = 86.79 × 10⁶ mm⁴

Answer

I_req = 86.79 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 160.2 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/2800² = 160173 N = 160.2 kN

Answer

P_cr = 160.2 kN

14.Column buckling

Answer: σ_cr = 7.896 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·5000 = 5000 mm

KL/r = 5000/10 = 500

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/500² = 7.896 MPa

Answer

σ_cr = 7.896 MPa

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