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
εT = 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

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. 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/ymax
Simply supported, uniform load
Mmax = wL²/8

Method

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid
ȳ = ΣAiyi/ΣAi
Rectangle
I = bh³/12
Parallel-axis theorem
I = Ic + A·d²
Composite section
I = Σ(Ic,i + Aidi²)

Mohr’s circle & stress transformation

Centre
C = (σx + σy)/2
Radius
R = √(((σx − σy)/2)² + τxy²)
Principal stresses
σ1,2 = C ± R
Principal direction
tan 2θp = 2τxy/(σx − σy)
Inclined plane
σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ

Method

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. 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
Pcr = π²EI/(KL)²
Effective length factor
K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress
σcr = Pcr/A
Slenderness
λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium
Ø 12 mmBrass2 mP = 30 kN

Find the elongation δ of the brass bar under the axial load.

P = 30 kNd = 12 mmL = 2 mE = 100 GPa
δ = mm
2.Axial stress & strainHard
Ø 35 mmBrass0.5 mØ 16 mmCopper0.5 mP = 60 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 60 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 0.5 m, copper (E = 117 GPa)
σmax = MPa
3.Torsion of shaftsMedium
d = 15 mmT = 100 N·m200 mmφ = 8.87°

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 100 N·m. The two ends of a gauge length of 200 mm rotate 8.87° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 100 N·md = 15 mmL = 200 mmφ = 8.87°
G = GPa
4.Torsion of shaftsHard
Tdo​ = 50 mm, di​ = 38 mm

A hollow shaft transmits 20 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.

P = 20 kWN = 500 rpmdo = 50 mmdi = 38 mm
τmax = MPa
5.Shear force & bending momentMedium
6 kN/m55 kNABCDEFG2 m2 m2 m2 m2 m2 m

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.

55 kN @ 8 m6 kN/m, 0–12 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
6.Shear force & bending momentHard
w = 12 kN/mP = 40 kNAB1 m3 m

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.

L = 4 mP = 40 kN at 1 mw = 12 kN/m from 0 to 4 m
Mmax = kN·m
7.Centroid & second moment of areaMedium
280150t = 10ȳ = ?dimensions in mm

A channel section lies with its open side up: a base plate 280 mm wide with two upright legs, 150 mm high overall. Base and legs are all 10 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 280 mmh = 150 mmt = 10 mm
ȳ = mm
8.Centroid & second moment of areaHard
17030022330228xdimensions in mm

A welded I-section has a top flange 170 × 22 mm, a web 330 mm high and 8 mm thick, and a bottom flange 300 × 22 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 170 × 22 mmweb: 8 × 330 mmbottom flange: 300 × 22 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 50σy​ = −50τxy​ = 60MPa

A point in a loaded part has σₓ = 50 MPa, σᵧ = −50 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 50 MPaσy = −50 MPaτxy = 60 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 120σy​ = 100τxy​ = 30MPa

A point in a loaded part has σₓ = 120 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σx = 120 MPaσy = 100 MPaτxy = 30 MPaθ = 40°
σθ = MPa
11.Beam deflectionMedium
w = 8 kN/mδL = 3 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.

w = 8 kN/mL = 3 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
P = 24 kNP = 24 kNABδ1.667 m1.666 m1.667 m

A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δmax.

P = 24 kNL = 5 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
PL = 3 md = 30 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
14.Column bucklingHard
P = 70 kNL = 1 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 70 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 70 kNn = 2L = 1 mE = 200 GPa (steel)
d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 5.305 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·12²/4 = 113.1 mm²
σ = P/A = 30·1000/113.1 = 265.3 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 265.3/100000 = 0.002653 = 2653 µε

Elongation

δ = ε·L = 0.002653·2000 mm = 5.305 mm

Answer

δ = 5.305 mm
2.Axial stress & strain

Answer: σmax = 298.4 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·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 62.36 MPa
σ2 = 298.4 MPa

Largest stress

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

σmax = 298.4 MPa

Answer

σmax = 298.4 MPa
3.Torsion of shafts

Answer: G = 25.99 GPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/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·15⁴/32 = 4970 mm⁴
φ = 8.87·π/180 = 0.1548 rad

Shear modulus

T = 100 N·m = 100000 N·mm
G = TL/(Jφ) = 100000·200/(4970·0.1548) = 25993 MPa
G = 25.99 GPa

Answer

G = 25.99 GPa

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

4.Torsion of shafts

Answer: τmax = 23.35 MPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 20000/52.36 = 382 N·m

Polar second moment of area

J = π(50⁴ − 38⁴)/32 = 408885 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 381972·25/408885 = 23.35 MPa

Answer

τmax = 23.35 MPa
5.Shear force & bending moment

Answer: E (x = 8 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 = 54.33 kN, R2 = 72.67 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 96.67, MC = 169.3, MD = 218, ME = 242.7 kN·m, MF = 133.3, MG = 0

Answer

Mark E: ME = 242.7 kN·m
6.Shear force & bending moment

Answer: Mmax = 48.17 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.

UDL resultant: W = w·ℓ = 12·4 = 48 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 40·1 + 12·4·2
RB = 136/4 = 34 kN
ΣFy = 0: RA = 88 − 34 = 54 kN

Where is V = 0?

RA − P − w·x0 = 0 → x0 = (54 − 40)/12 = 1.167 m

Maximum moment

Mmax = RA·x0 − P·(x0 − 1) − w·x0²/2 = 48.17 kN·m

Answer

Mmax = 48.17 kN·m

It occurs at x = 1.167 m from A.

7.Centroid & second moment of area

Answer: ȳ = 42.5 mm

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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A1 = 280·10 = 2800 mm², y1 = 5 mm
two legs: A2 = 2·10·140 = 2800 mm², y2 = 10 + 140/2 = 80 mm

Centroid

ȳ = (2800·5 + 2800·80)/5600 = 42.5 mm

Answer

ȳ = 42.5 mm
8.Centroid & second moment of area

Answer: Ix = 325.1 × 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 = 3740 mm², y = 363 mm
web: A = 2640 mm², y = 187 mm
bottom flange: A = 6600 mm², y = 11 mm

Centroid

ȳ = (3740·363 + 2640·187 + 6600·11)/12980 = 148.2 mm

Parallel-axis theorem

top flange: 170·22³/12 + 3740·214.8² = 1.727 × 10⁸ mm⁴
web: 8·330³/12 + 2640·38.78² = 2.793 × 10⁷ mm⁴
bottom flange: 300·22³/12 + 6600·137.2² = 1.245 × 10⁸ mm⁴
Ix = 1.727 × 10⁸ + 2.793 × 10⁷ + 1.245 × 10⁸
Ix = 3.251 × 10⁸ mm⁴ = 325.1 × 10⁶ mm⁴

Answer

Ix = 325.1 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τmax = 78.1 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((50 − −50)/2)² + 60²) = 78.1 MPa

Answer

τmax = 78.1 MPa
10.Mohr’s circle & stress transformation

Answer: σθ = 141.3 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

σθ = 110 + 10·0.1736 + 30·0.9848 = 141.3 MPa

Answer

σθ = 141.3 MPa
11.Beam deflection

Answer: δmax = 3.348 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 = 8 N/mm, L = 3000 mm
EI = 2.52 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·8·3000⁴/(384·2.52 × 10¹²) = 3.348 mm

Answer

δmax = 3.348 mm
12.Beam deflection

Answer: δmax = 9.34 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 = 24000 N, L = 5000 mm
EI = 1.14 × 10¹³ N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 1667 mm
δ₁ = 4.67 mm

Add both loads

δmax = 2·δ₁ = 23PL³/(648EI) = 9.34 mm

Answer

δmax = 9.34 mm
13.Column buckling

Answer: Pcr = 2.18 kN

Euler buckling: a slender column becomes unstable at Pcr = 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/6000² = 2180 N = 2.18 kN

Answer

Pcr = 2.18 kN
14.Column buckling

Answer: d = 34.67 mm

Euler buckling: a slender column becomes unstable at Pcr = 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.

Required critical load

Pcr = n·P = 2·70000 = 140000 N
KL = 1·1000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·140000·1000²/(π³·200000))^(1/4) = 34.67 mm

Answer

d = 34.67 mm

Check: the slenderness KL/r is 115.4, so the column is slender and Euler applies.

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