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 mmAluminium3 mP = 55 kN

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

P = 55 kNd = 20 mmL = 3 mE = 70 GPa
δ = mm
2.Axial stress & strainHard
Ø 30 mmBrass1 mØ 12 mmSteel1 mP = 115 kN

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

P = 115 kNsegment 1: Ø30 mm, 1 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)
δ = mm
3.Torsion of shaftsMedium
d = ?T = 250 N·mL = 1 m

A solid aluminium shaft 1 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 250 N·mL = 1 mG = 26 GPaφallow = 0.5°
d = mm
4.Torsion of shaftsHard
d = 50 mmT0​ = 1000 N·mACBa = 1 mb = 2 m

A solid aluminium shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 1 m from A and 2 m from B. Find the angle of rotation φC of section C, in degrees.

d = 50 mmT0 = 1000 N·ma = 1 mb = 2 mG = 26 GPa
φC = °
5.Shear force & bending momentMedium
w = 12 kN/mABsection0.5 m5.5 m

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

L = 6 mw = 12 kN/m from 0 to 6 mx = 0.5 m
Mx = kN·m
6.Shear force & bending momentHard
w = 12 kN/mP = 20 kNABsection1 m1.5 m1.5 m

Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 20 kN at 2.5 mw = 12 kN/m from 0 to 4 mx = 1 m
Mx = kN·m
7.Centroid & second moment of areaMedium
160151808xdimensions in mm

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

bf = 160 mmtf = 15 mmtw = 8 mmhw = 180 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
120 mm300 mm195 mm30 × 40xIx​ = ?

A 120 mm × 300 mm plate has a rectangular cut-out 30 mm wide and 40 mm high, centred on the vertical centre line with its centre 195 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 120 mmh = 300 mmcut-out: 30 × 40 mmhole centre: y = 195 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 100σy​ = −20τxy​ = 30MPa

A point in a loaded part has σₓ = 100 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 100 MPaσy = −20 MPaτxy = 30 MPa
θp = °
10.Mohr’s circle & stress transformationHard
σx​ = 0σy​ = −50τxy​ = 40MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = −50 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σx = 0 MPaσy = −50 MPaτxy = 40 MPaθ = 20°
σθ = MPa
11.Beam deflectionMedium
w = 2 kN/mδL = 4 m

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

w = 2 kN/mL = 4 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
w = 1 kN/mABδ4 m

A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/300. Find the smallest depth hreq of the section.

w = 1 kN/mL = 4 mE = 11 GPa (timber)b = 125 mm
hreq = mm
13.Column bucklingMedium
PL = 6 md = 50 mmcross-section

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

L = 6 mE = 70 GPa (aluminium)d = 50 mm
Pcr = kN
14.Column bucklingHard
P = 120 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 120 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 120 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 7.503 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 = 55·1000/314.2 = 175.1 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 175.1/70000 = 0.002501 = 2501 µε

Elongation

δ = ε·L = 0.002501·3000 mm = 7.503 mm

Answer

δ = 7.503 mm
2.Axial stress & strain

Answer: δ = 6.711 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·12²/4 = 113.1 mm²
σ1 = 162.7 MPa
σ2 = 1017 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 162.7·1000/100000 = 1.627 mm
δ2 = σ2L2/E2 = 1017·1000/200000 = 5.084 mm
δ = δ1 + δ2 = 6.711 mm

Answer

δ = 6.711 mm
3.Torsion of shafts

Answer: d = 57.88 mm

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.

Required polar moment

T = 250 N·m = 250000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 0.5·π/180 = 0.008727 rad
J = TL/(Gφ) = 250000·1000/(26000·0.008727) = 1.102 × 10⁶ mm⁴

Diameter

d = (32J/π)1/4 = 57.88 mm

Answer

d = 57.88 mm

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

4.Torsion of shafts

Answer: φC = 2.394 °

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 1000
TA·a = TB·b
TA = 1000·2/3 = 666.7 N·m
TB = 1000·1/3 = 333.3 N·m

Rotation of C

Use segment AC: it is fixed at A and carries TA.

J = π·50⁴/32 = 613592 mm⁴
G = 26 GPa = 26000 MPa
φC = TA·a/(GJ) = 666667·1000/(26000·613592) = 0.04179 rad = 2.394°

Answer

φC = 2.394°
5.Shear force & bending moment

Answer: Mx = 16.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·6 = 72 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 12·6·3
RB = 216/6 = 36 kN
ΣFy = 0: RA = 72 − 36 = 36 kN
w = 12 kN/mABRA​ = 36RB​ = 36section0.5 m5.5 m

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.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 36·0.5 − 12·0.5·0.25 = 16.5 kN·m

Answer

Mx = 16.5 kN·m
6.Shear force & bending moment

Answer: Mx = 25.5 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.

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 = 20·2.5 + 12·4·2
RB = 146/4 = 36.5 kN
ΣFy = 0: RA = 68 − 36.5 = 31.5 kN
w = 12 kN/mP = 20 kNABRA​ = 31.5RB​ = 36.5section1 m1.5 m1.5 m

Cut at x = 1 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 = RA·x − (each load left of the cut × its distance to the cut)
Mx = 31.5·1 − 12·1·0.5 = 25.5 kN·m

Answer

Mx = 25.5 kN·m
7.Centroid & second moment of area

Answer: Ix = 12.49 × 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 = 160·15 = 2400 mm², yf = 180 + 15/2 = 187.5 mm
web: Aw = 8·180 = 1440 mm², yw = 180/2 = 90 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (2400·187.5 + 1440·90)/3840 = 150.9 mm

Parallel-axis theorem

flange: 160·15³/12 + 2400·(187.5 − 150.9)² = 3.253 × 10⁶ mm⁴
web: 8·180³/12 + 1440·(90 − 150.9)² = 9.235 × 10⁶ mm⁴
Ix = 3.253 × 10⁶ + 9.235 × 10⁶ = 1.249 × 10⁷ mm⁴ = 12.49 × 10⁶ mm⁴

Answer

Ix = 12.49 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: Ix = 267.3 × 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 = (36000·150 − 1200·195)/(36000 − 1200) = 148.4 mm

Full plate about ȳ

120·300³/12 + 36000·1.552² = 270.1 × 10⁶ mm⁴

Hole about ȳ

30·40³/12 + 1200·46.55² = 2.76 × 10⁶ mm⁴

Subtract

Ix = 267.3 × 10⁶ mm⁴

Answer

Ix = 267.3 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: θp = 13.28 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·30, 100 − −20) = 26.57°

Angle on the element

θp = 13.28°

Answer

θp = 13.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σθ = 19.86 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 = −25 + 25·0.766 + 40·0.6428 = 19.86 MPa

Answer

σθ = 19.86 MPa
11.Beam deflection

Answer: δmax = 1.134 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 = 2 N/mm, L = 4000 mm
EI = 5.88 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·2·4000⁴/(384·5.88 × 10¹²) = 1.134 mm

Answer

δmax = 1.134 mm
12.Beam deflection

Answer: hreq = 129.7 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 = 1 N/mm, L = 4000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/300 = 4000/300 = 13.33 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·1·4000⁴/(384·11000·13.33) = 2.273 × 10⁷ mm⁴

Required depth

bh³/12 = Ireq → h = (12·2.273 × 10⁷/125)1/3 = 129.7 mm

Answer

hreq = 129.7 mm

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

13.Column buckling

Answer: Pcr = 23.55 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·306796/3000² = 23551 N = 23.55 kN

Answer

Pcr = 23.55 kN
14.Column buckling

Answer: d = 102.7 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 = 2.5·120000 = 300000 N
KL = 2·3000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·6000²/(π³·200000))^(1/4) = 102.7 mm

Answer

d = 102.7 mm

Check: the slenderness KL/r is 233.6, so the column is slender and Euler applies.

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