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 10 mmAluminium1.5 mP = 60 kN

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

P = 60 kNd = 10 mmL = 1.5 mE = 70 GPa
ε = µε
2.Axial stress & strainHard
Ø 25 mmCopper2 mØ 16 mmAluminium1.5 mP = 60 kN

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

P = 60 kNsegment 1: Ø25 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
3.Torsion of shaftsMedium
d = ?T = 200 N·mL = 1.5 m

A solid aluminium shaft 1.5 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.

T = 200 N·mL = 1.5 mG = 26 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
do​ = ?75 kW, 3000 rpm

A hollow shaft must transmit 75 kW at 3000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.

P = 75 kWN = 3000 rpmdi/do = 0.6τallow = 60 MPa
do = mm
5.Shear force & bending momentMedium
w = 10 kN/mABsection6 m4 m

Find the bending moment Mx at the section x = 6 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 10 kN/m from 0 to 10 mx = 6 m
Mx = kN·m
6.Shear force & bending momentHard
4 kN/m60 kN60 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.

60 kN @ 12 m60 kN @ 8 m4 kN/m, 0–10 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)
7.Centroid & second moment of areaMedium
1302201814xdimensions in mm

A symmetric I-section has flanges 130 mm wide and 18 mm thick, an overall depth of 220 mm and a web 14 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.

B = 130 mmH = 220 mmtf = 18 mmtw = 14 mm
Sx = × 10³ mm³
8.Centroid & second moment of areaHard
1203030012xsagging Mdimensions in mm

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 35 MPa in tension and 150 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.

bf = 120 mmtf = 30 mmtw = 12 mmhw = 300 mmσt,allow = 35 MPaσc,allow = 150 MPa
Mallow = kN·m
9.Mohr’s circle & stress transformationMedium
σx​ = −30MPaτxy​ = ?

A point has σₓ = −30 MPa and σᵧ = 0 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 20 MPa?

σx = −30 MPaσy = 0 MPaσ1 = 20 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 10σy​ = −60τxy​ = 30MPa

A point in a loaded part has σₓ = 10 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.

σx = 10 MPaσy = −60 MPaτxy = 30 MPaθ = 30°
σθ = MPa
11.Beam deflectionMedium
w = 5 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 = 5 kN/mL = 3 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴
δ = mm
12.Beam deflectionHard
w = 2 kN/mABδ5 m

A simply supported timber beam of rectangular section, 150 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth hreq of the section.

w = 2 kN/mL = 5 mE = 11 GPa (timber)b = 150 mm
hreq = mm
13.Column bucklingMedium
PL = 2.5 m20 × 80 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)b × h = 20 × 80 mm
Pcr = kN
14.Column bucklingHard
P = 65 kNL = 1.5 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 65 kNn = 2.5L = 1.5 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 10913 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·10²/4 = 78.54 mm²
σ = P/A = 60·1000/78.54 = 763.9 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 763.9/70000 = 0.01091 = 10913 µε

Answer

ε = 10913 µε
2.Axial stress & strain

Answer: δ = 8.484 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 = 122.2 MPa
σ2 = 298.4 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 122.2·2000/117000 = 2.089 mm
δ2 = σ2L2/E2 = 298.4·1500/70000 = 6.395 mm
δ = δ1 + δ2 = 8.484 mm

Answer

δ = 8.484 mm
3.Torsion of shafts

Answer: d = 46.03 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 = 200 N·m = 200000 N·mm
L = 1.5 m = 1500 mm
G = 26 GPa = 26000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 200000·1500/(26000·0.02618) = 440737 mm⁴

Diameter

d = (32J/π)1/4 = 46.03 mm

Answer

d = 46.03 mm

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

4.Torsion of shafts

Answer: do = 28.55 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.

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s
T = P/ω = 75000/314.2 = 238.7 N·m

Outer diameter from the allowable stress

T = 238.7 N·m = 238732 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·238732/(π·60·0.8704)) = 28.55 mm
di = 0.6·28.55 = 17.13 mm

Answer

do = 28.55 mm

A solid shaft would need d = 27.26 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: Mx = 120 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 = 10·10 = 100 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 10·10·5
RB = 500/10 = 50 kN
ΣFy = 0: RA = 100 − 50 = 50 kN
w = 10 kN/mABRA​ = 50RB​ = 50section6 m4 m

Cut at x = 6 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 = 50·6 − 10·6·3 = 120 kN·m

Answer

Mx = 120 kN·m
6.Shear force & bending moment

Answer: F (x = 10 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 = 20 kN, R2 = 140 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 = 32, MC = 48, MD = 48, ME = 32, MF = −120 kN·m, MG = 0

Answer

Mark F: MF = −120 kN·m
7.Centroid & second moment of area

Answer: Sx = 501.2 × 10³ mm³

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 = Ic + 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 − tw) wide and hw = H − 2tf high, and share the axis x.

hw = 220 − 2·18 = 184 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (130·220³ − 116·184³)/12
Ix = 5.513 × 10⁷ mm⁴

Section modulus

c = H/2 = 110 mm
Sx = I/c = 5.513 × 10⁷/110
Sx = 501225 mm³ = 501.2 × 10³ mm³

Answer

Sx = 501.2 × 10³ mm³
8.Centroid & second moment of area

Answer: Mallow = 11.48 kN·m

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.

Centroid

ȳ = (3600·315 + 3600·150)/7200 = 232.5 mm
top edge: c = 330 − 232.5 = 97.5 mm

Second moment of area

I = 120·30³/12 + 3600·82.5² + 12·300³/12 + 3600·82.5² = 7.628 × 10⁷ mm⁴

Moment allowed by each edge

Sagging: the top is in compression, the bottom in tension.

top edge: M = 150·7.628 × 10⁷/97.5 = 117.3 kN·m
bottom edge: M = 35·7.628 × 10⁷/232.5 = 11.48 kN·m

Governing value

Mallow = min = 11.48 kN·m (bottom edge)

Answer

Mallow = 11.48 kN·m
9.Mohr’s circle & stress transformation

Answer: τxy = 31.62 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 = (−30 + 0)/2 = −15 MPa
R = σ1 − C = 20 − (−15) = 35 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(35² − 15²) = 31.62 MPa

Answer

τxy = 31.62 MPa
10.Mohr’s circle & stress transformation

Answer: σθ = 18.48 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 + 35·0.5 + 30·0.866 = 18.48 MPa

Answer

σθ = 18.48 MPa
11.Beam deflection

Answer: δ = 54.79 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 = 5 N/mm, L = 3000 mm
EI = 9.24 × 10¹¹ N·mm²

Deflection

δ = wL⁴/(8EI) = 5·3000⁴/(8·9.24 × 10¹¹) = 54.79 mm

Answer

δ = 54.79 mm
12.Beam deflection

Answer: hreq = 204.3 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 = 5000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·2·5000⁴/(384·11000·13.89) = 1.065 × 10⁸ mm⁴

Required depth

bh³/12 = Ireq → h = (12·1.065 × 10⁸/150)1/3 = 204.3 mm

Answer

hreq = 204.3 mm

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

13.Column buckling

Answer: Pcr = 67.38 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

Imin = h·b³/12 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·53333/1250² = 67376 N = 67.38 kN

Answer

Pcr = 67.38 kN
14.Column buckling

Answer: d = 81.04 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 = 2.5·65000 = 162500 N
KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·3000²/(π³·70000))^(1/4) = 81.04 mm

Answer

d = 81.04 mm

Check: the slenderness KL/r is 148.1, so the column is slender and Euler applies.

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