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 20 mmCopper1.5 mP = 70 kN

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 70 kNd = 20 mmL = 1.5 mE = 117 GPa
ε = µε
2.Axial stress & strainHard
Ø 30 mmBrass1.5 mØ 16 mmSteel1 mP = 40 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 40 kNsegment 1: Ø30 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, steel (E = 200 GPa)
δ = mm
3.Torsion of shaftsMedium
T = 250 N·md = 25 mmL = 0.5 mφ = ?

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 = 250 N·md = 25 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 40 mm, di​ = 28 mm

A hollow shaft transmits 50 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.

P = 50 kWN = 1450 rpmdo = 40 mmdi = 28 mm
τmax = MPa
5.Shear force & bending momentMedium
w = 4 kN/mP = 50 kNABsection2 m0.5 m0.5 m

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2.5 m from A. Find the bending moment Mx at the section x = 2 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 3 mP = 50 kN at 2.5 mw = 4 kN/m from 0 to 3 mx = 2 m
Mx = kN·m
6.Shear force & bending momentHard
20 kN30 kNABCDEFG1 m1 m1 m1 m1 m1 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.

20 kN @ 6 m30 kN @ 4 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
7.Centroid & second moment of areaMedium
1201528014xdimensions in mm

Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.

bf = 120 mmtf = 15 mmtw = 14 mmhw = 280 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
150 mm200 mm80 mm50 × 20xIx​ = ?

A 150 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 80 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 150 mmh = 200 mmcut-out: 50 × 20 mmhole centre: y = 80 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 50MPaτxy​ = ?

A point has σₓ = 110 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 145 MPa?

σx = 110 MPaσy = 50 MPaσ1 = 145 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 110σy​ = 40τxy​ = −40MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = 40 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σx = 110 MPaσy = 40 MPaτxy = −40 MPaθ = 35°
σθ = MPa
11.Beam deflectionMedium
w = 8 kN/mδL = 3 m

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 8 kN/mL = 3 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴
δ = mm
12.Beam deflectionHard
P = 8 kNABδ3.5 m1.5 m

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the maximum deflection δmax of the beam.

P = 8 kNb = 1.5 mL = 5 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
PL = 2.5 md = 30 mmcross-section

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

L = 2.5 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
14.Column bucklingHard
P = 105 kNL = 2 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 105 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 105 kNn = 3L = 2 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 1904 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·20²/4 = 314.2 mm²
σ = P/A = 70·1000/314.2 = 222.8 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 222.8/117000 = 0.001904 = 1904 µε

Answer

ε = 1904 µε
2.Axial stress & strain

Answer: δ = 1.844 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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 56.59 MPa
σ2 = 198.9 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 56.59·1500/100000 = 0.8488 mm
δ2 = σ2L2/E2 = 198.9·1000/200000 = 0.9947 mm
δ = δ1 + δ2 = 1.844 mm

Answer

δ = 1.844 mm
3.Torsion of shafts

Answer: φ = 2.334 °

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 second moment of area

J = π·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·500/(80000·38350) = 0.04074 rad
φ = 0.04074·180/π = 2.334°

Answer

φ = 2.334°
4.Torsion of shafts

Answer: τmax = 34.48 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·1450/60 = 151.8 rad/s
T = P/ω = 50000/151.8 = 329.3 N·m

Polar second moment of area

J = π(40⁴ − 28⁴)/32 = 190984 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 329286·20/190984 = 34.48 MPa

Answer

τmax = 34.48 MPa
5.Shear force & bending moment

Answer: Mx = −27 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.

Keep the free-end part

P: −50·0.5 = −25 kN·m
w: −4·1²/2 = −2 kN·m
Mx = −25 − 2 = −27 kN·m

Answer

Mx = −27 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: F (x = 5 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 = 2 kN, R2 = 48 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 = 2, MC = 4, MD = 6, ME = 8, MF = −20 kN·m, MG = 0

Answer

Mark F: MF = −20 kN·m
7.Centroid & second moment of area

Answer: Ix = 52.48 × 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 two rectangles

Measure every y from the bottom edge.

flange: Af = 120·15 = 1800 mm², yf = 280 + 15/2 = 287.5 mm
web: Aw = 14·280 = 3920 mm², yw = 280/2 = 140 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (1800·287.5 + 3920·140)/5720 = 186.4 mm

Parallel-axis theorem

flange: 120·15³/12 + 1800·(287.5 − 186.4)² = 1.843 × 10⁷ mm⁴
web: 14·280³/12 + 3920·(140 − 186.4)² = 3.406 × 10⁷ mm⁴
Ix = 1.843 × 10⁷ + 3.406 × 10⁷ = 5.248 × 10⁷ mm⁴ = 52.48 × 10⁶ mm⁴

Answer

Ix = 52.48 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: Ix = 99.55 × 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 = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.

Centroid

ȳ = (30000·100 − 1000·80)/(30000 − 1000) = 100.7 mm

Full plate about ȳ

150·200³/12 + 30000·−0.6897² = 100 × 10⁶ mm⁴

Hole about ȳ

50·20³/12 + 1000·−20.69² = 0.4614 × 10⁶ mm⁴

Subtract

Ix = 99.55 × 10⁶ mm⁴

Answer

Ix = 99.55 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τxy = 57.66 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 = (110 + 50)/2 = 80 MPa
R = σ1 − C = 145 − 80 = 65 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 30²) = 57.66 MPa

Answer

τxy = 57.66 MPa
10.Mohr’s circle & stress transformation

Answer: σθ = 49.38 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 = 75 + 35·0.342 + −40·0.9397 = 49.38 MPa

Answer

σθ = 49.38 MPa
11.Beam deflection

Answer: δ = 11.25 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 = 7.2 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 8·3000⁴/(8·7.2 × 10¹²) = 11.25 mm

Answer

δ = 11.25 mm
12.Beam deflection

Answer: δmax = 18.08 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 = 8000 N, b = 1500 mm, L = 5000 mm
EI = 9.24 × 10¹¹ N·mm²

Position of the maximum

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

x = √((L² − b²)/3) = √((5000² − 1500²)/3) = 2754 mm

Maximum deflection

δmax = Pb(L² − b²)3/2/(9√3·EIL) = 18.08 mm

Answer

δmax = 18.08 mm
13.Column buckling

Answer: Pcr = 50.23 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 = 0.5·2500 = 1250 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/1250² = 50230 N = 50.23 kN

Answer

Pcr = 50.23 kN
14.Column buckling

Answer: d = 78.07 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 = 3·105000 = 315000 N
KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·315000·2000²/(π³·70000))^(1/4) = 78.07 mm

Answer

d = 78.07 mm

Check: the slenderness KL/r is 102.5, so the column is slender and Euler applies.

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