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

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 70 GPaL = 500 mm

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 = 35 kNsegment 1: Ø25 mm, 1 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 400 N·md = 30 mmL = 0.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 1000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 20 kWN = 1000 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 3 kN/m from 0 to 8 mx = 0.5 m

V_x = kN

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1 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 = 5 mw = 6 kN/m from 0 to 5 msupports at x = 0 and x = 4 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 200 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 70 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 200 mmd = 60 mmhole centre: y = 70 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 130 × 14 mm, a web 160 mm high and 16 mm thick, and a bottom flange 170 × 10 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 130 × 14 mmweb: 16 × 160 mmbottom flange: 170 × 10 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −40 MPa, σᵧ = 10 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −40 MPaσ_y = 10 MPaτ_xy = 10 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 150 MPa, σᵧ = 10 MPa and τₓᵧ = −20 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 = 150 MPaσ_y = 10 MPaτ_xy = −20 MPaθ = 45°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2 m from A. Find the deflection δ_B at the free end B.

P = 7 kNa = 2 mL = 2.5 mE = 70 GPa (aluminium)I = 23 × 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 = 8 kN/mL = 2.5 mE = 11 GPa (timber)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)b × h = 40 × 100 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 95 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 95 kNn = 3L = 1.5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.3701 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·2199 = 4.053 × 10⁸ N

Shortening

δ = 300000·500/4.053 × 10⁸ = 0.3701 mm

Answer

δ = 0.3701 mm

2.Axial stress & strain

Answer: δ = 4.443 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 = 71.3 MPa

σ₂ = 174.1 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 71.3·1000/100000 = 0.713 mm

δ₂ = σ₂L₂/E₂ = 174.1·1500/70000 = 3.73 mm

δ = δ₁ + δ₂ = 4.443 mm

Answer

δ = 4.443 mm

3.Torsion of shafts

Answer: φ = 1.801 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·500/(80000·79522) = 0.03144 rad

φ = 0.03144·180/π = 1.801°

Answer

φ = 1.801°

4.Torsion of shafts

Answer: d = 25.31 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 20000/104.7 = 191 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·190986/(π·60)) = 25.31 mm

Answer

d = 25.31 mm

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

5.Shear force & bending moment

Answer: V_x = 10.5 kN

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·8 = 24 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 3·8·4

R_B = 96/8 = 12 kN

ΣF_y = 0: R_A = 24 − 12 = 12 kN

Cut at x = 0.5 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.

V = R_A − (all downward loads left of the cut)

V_x = 12 − 3·0.5 = 10.5 kN

Answer

V_x = 10.5 kN

6.Shear force & bending moment

Answer: M_max = 10.55 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 = 6·5 = 30 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·4 = 6·5·2.5

R_B = 75/4 = 18.75 kN

ΣF_y = 0: R_A = 30 − 18.75 = 11.25 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 11.25/6 = 1.875 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 11.25²/(2·6) = 10.55 kN·m

Answer

M_max = 10.55 kN·m

7.Centroid & second moment of area

Answer: I_x = 63.07 × 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 = (20000·100 − 2827·70)/(20000 − 2827) = 104.9 mm

Full plate about ȳ

100·200³/12 + 20000·−4.939² = 67.15 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·−34.94² = 4.088 × 10⁶ mm⁴

Subtract

I_x = 63.07 × 10⁶ mm⁴

Answer

I_x = 63.07 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 31.53 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 1820 mm², y = 177 mm

web: A = 2560 mm², y = 90 mm

bottom flange: A = 1700 mm², y = 5 mm

Centroid

ȳ = (1820·177 + 2560·90 + 1700·5)/6080 = 92.28 mm

Parallel-axis theorem

top flange: 130·14³/12 + 1820·84.72² = 1.309 × 10⁷ mm⁴

web: 16·160³/12 + 2560·2.276² = 5.475 × 10⁶ mm⁴

bottom flange: 170·10³/12 + 1700·87.28² = 1.296 × 10⁷ mm⁴

I_x = 1.309 × 10⁷ + 5.475 × 10⁶ + 1.296 × 10⁷

I_x = 3.153 × 10⁷ mm⁴ = 31.53 × 10⁶ mm⁴

Answer

I_x = 31.53 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 26.93 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 = √(((−40 − 10)/2)² + 10²) = 26.93 MPa

Answer

τ_max = 26.93 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 60 MPa

Transformation equation

σ_θ = 80 + 70·0 + −20·1 = 60 MPa

Answer

σ_θ = 60 MPa

11.Beam deflection

Answer: δ_B = 15.94 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 = 7000 N, a = 2000 mm, L = 2500 mm

EI = 1.61 × 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) = 11.59 mm

θ_P = Pa²/(2EI) = 0.008696 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 15.94 mm

Answer

δ_B = 15.94 mm

12.Beam deflection

Answer: I_req = 53.27 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 2500/360 = 6.944 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·8·2500⁴/(384·11000·6.944) = 5.327 × 10⁷ mm⁴

I_req = 53.27 × 10⁶ mm⁴

Answer

I_req = 53.27 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 168.4 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/2500² = 168441 N = 168.4 kN

Answer

P_cr = 168.4 kN

14.Column buckling

Answer: d = 93.26 mm

Required critical load

P_cr = n·P = 3·95000 = 285000 N

KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·285000·3000²/(π³·70000))^(1/4) = 93.26 mm

Answer

d = 93.26 mm

Check: the slenderness KL/r is 128.7, so the column is slender and Euler applies.

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