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 = 40 kNd = 25 mmL = 3 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 = 105 kNsegment 1: Ø35 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 0.5 m, brass (E = 100 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel shaft 1 m long carries a torque of 2000 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 2000 N·mL = 1 mG = 80 GPaφ_allow = 2°

d = mm

4.Torsion of shaftsHard

A solid steel shaft transmits 5 kW at 1000 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 1000 rpmτ_allow = 80 MPaθ_allow = 0.75°/mG = 80 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 5.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 10 kN/m from 0 to 6 mx = 5.5 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 2 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 = 10 mw = 3 kN/m from 0 to 10 msupports at x = 0 and x = 8 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 220 mm wide and 16 mm thick, an overall depth of 400 mm and a web 10 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 220 mmH = 400 mmt_f = 16 mmt_w = 10 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 150 × 18 mm, a web 400 mm high and 8 mm thick, and a bottom flange 210 × 10 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 150 × 18 mmweb: 8 × 400 mmbottom flange: 210 × 10 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 90 MPa and σᵧ = 100 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 145 MPa?

σ_x = 90 MPaσ_y = 100 MPaσ₁ = 145 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 150 MPa, σᵧ = 60 MPa and τₓᵧ = −30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 150 MPaσ_y = 60 MPaτ_xy = −30 MPaθ = -20°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 10 kN/mL = 8 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 1 kN/mL = 4 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

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

L = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel 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 = 3 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 696.5 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·25²/4 = 490.9 mm²

σ = P/A = 40·1000/490.9 = 81.49 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 81.49/117000 = 0.0006965 = 696.5 µε

Answer

ε = 696.5 µε

2.Axial stress & strain

Answer: δ = 2.884 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·16²/4 = 201.1 mm²

σ₁ = 109.1 MPa

σ₂ = 522.2 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 109.1·500/200000 = 0.2728 mm

δ₂ = σ₂L₂/E₂ = 522.2·500/100000 = 2.611 mm

δ = δ₁ + δ₂ = 2.884 mm

Answer

δ = 2.884 mm

3.Torsion of shafts

Answer: d = 51.97 mm

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.

Required polar moment

T = 2000 N·m = 2 × 10⁶ N·mm

L = 1 m = 1000 mm

G = 80 GPa = 80000 MPa

φ = 2·π/180 = 0.03491 rad

J = TL/(Gφ) = 2 × 10⁶·1000/(80000·0.03491) = 716197 mm⁴

Diameter

d = (32J/π)^1/4 = 51.97 mm

Answer

d = 51.97 mm

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

4.Torsion of shafts

Answer: d = 26.11 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 5000/104.7 = 47.75 N·m

Diameter for strength

T = 47.75 N·m = 47746 N·mm

d_τ = ∛(16·47746/(π·80)) = 14.49 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 80 GPa = 80000 MPa

d_φ = (32·47746/(π·80000·1.309 × 10⁻⁵))^1/4 = 26.11 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(14.49, 26.11) = 26.11 mm

Answer

d = 26.11 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 13.75 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·6 = 60 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 10·6·3

R_B = 180/6 = 30 kN

ΣF_y = 0: R_A = 60 − 30 = 30 kN

Cut at x = 5.5 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.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 30·5.5 − 10·5.5·2.75 = 13.75 kN·m

Answer

M_x = 13.75 kN·m

6.Shear force & bending moment

Answer: M_max = 21.09 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 = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·8 = 3·10·5

R_B = 150/8 = 18.75 kN

ΣF_y = 0: R_A = 30 − 18.75 = 11.25 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 11.25/3 = 3.75 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 11.25²/(2·3) = 21.09 kN·m

Answer

M_max = 21.09 kN·m

7.Centroid & second moment of area

Answer: I_x = 301.2 × 10⁶ mm⁴

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.

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 = 400 − 2·16 = 368 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (220·400³ − 210·368³)/12

I_x = 3.012 × 10⁸ mm⁴ = 301.2 × 10⁶ mm⁴

Answer

I_x = 301.2 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 246.7 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 2700 mm², y = 419 mm

web: A = 3200 mm², y = 210 mm

bottom flange: A = 2100 mm², y = 5 mm

Centroid

ȳ = (2700·419 + 3200·210 + 2100·5)/8000 = 226.7 mm

Parallel-axis theorem

top flange: 150·18³/12 + 2700·192.3² = 9.989 × 10⁷ mm⁴

web: 8·400³/12 + 3200·16.72² = 4.356 × 10⁷ mm⁴

bottom flange: 210·10³/12 + 2100·221.7² = 1.033 × 10⁸ mm⁴

I_x = 9.989 × 10⁷ + 4.356 × 10⁷ + 1.033 × 10⁸

I_x = 2.467 × 10⁸ mm⁴ = 246.7 × 10⁶ mm⁴

Answer

I_x = 246.7 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 49.75 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 = (90 + 100)/2 = 95 MPa

R = σ₁ − C = 145 − 95 = 50 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 5²) = 49.75 MPa

Answer

τ_xy = 49.75 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 5.944 MPa

Transformation equation

τ_θ = −45·−0.6428 + −30·0.766 = 5.944 MPa

Answer

τ_θ = 5.944 MPa

11.Beam deflection

Answer: θ_A = 0.02963 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 8000 mm

EI = 7.2 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 10·8000³/(24·7.2 × 10¹²) = 0.02963 rad

Answer

θ_A = 0.02963 rad

That is 1.698°.

12.Beam deflection

Answer: h_req = 148.5 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.

w = 1 N/mm, L = 4000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1·4000⁴/(384·11000·11.11) = 2.727 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·2.727 × 10⁷/100)^1/3 = 148.5 mm

Answer

h_req = 148.5 mm

In practice, choose the next standard depth above this value.

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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/8000² = 21704 N = 21.7 kN

Answer

P_cr = 21.7 kN

14.Column buckling

Answer: σ_cr = 11.19 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.7·3000 = 2100 mm

KL/r = 2100/5 = 420

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/420² = 11.19 MPa

Answer

σ_cr = 11.19 MPa

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