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 30 mmAluminium1 mP = 115 kN

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

P = 115 kNd = 30 mmL = 1 mE = 70 GPa
δ = mm
2.Axial stress & strainHard
60 kN12500 mm500 mmØ25Ø50σ2​ = ?

A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 50 mm, length 500 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: copper, Ø50 mm, 500 mm, E = 120 GPaP = 60 kN
σ2 = MPa
3.Torsion of shaftsMedium
do​ = 80, di​ = 40 mmT = 1500 N·mL = 3 m

A hollow aluminium shaft (outer diameter 80 mm, inner diameter 40 mm, length 3 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.

T = 1500 N·mdo = 80 mmdi = 40 mmL = 3 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
d = 60 mmTB​ = 400 N·mTC​ = 750 N·mABC1.5 m2.5 m

A solid steel shaft of diameter 60 mm is fixed at A. A torque of 400 N·m acts at B, 1.5 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the magnitude of the angle of twist φC of end C, in degrees.

d = 60 mmTB = 400 N·mTC = 750 N·mAB = 1.5 mBC = 2.5 mG = 80 GPa
φC = °
5.Shear force & bending momentMedium
60 kN10 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 @ 4 m10 kN @ 6 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 = 5 kN/mABC4 m1.5 m

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment Mmax between the supports.

L = 5.5 mw = 5 kN/m from 0 to 5.5 msupports at x = 0 and x = 4 m
Mmax = kN·m
7.Centroid & second moment of areaMedium
3002528010xdimensions in mm

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

bf = 300 mmtf = 25 mmtw = 10 mmhw = 280 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
190300101602014xdimensions in mm

A welded I-section has a top flange 190 × 10 mm, a web 160 mm high and 14 mm thick, and a bottom flange 300 × 20 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 190 × 10 mmweb: 14 × 160 mmbottom flange: 300 × 20 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 40σy​ = −30τxy​ = 60MPa

A point in a loaded part has σₓ = 40 MPa, σᵧ = −30 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 40 MPaσy = −30 MPaτxy = 60 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 0σy​ = 80τxy​ = 40MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = 80 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σx = 0 MPaσy = 80 MPaτxy = 40 MPaθ = -20°
τθ = 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 = 11 GPa (timber)I = 23 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
w = 8 kN/mδL = 4 m

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.

w = 8 kN/mL = 4 mE = 11 GPa (timber)
Ireq = × 10⁶ mm⁴
13.Column bucklingMedium
PL = 1.5 m40 × 100 mmcross-section

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

L = 1.5 mE = 70 GPa (aluminium)b × h = 40 × 100 mm
Pcr = kN
14.Column bucklingHard
P = 90 kNL = 3 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 90 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 90 kNn = 3L = 3 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 2.324 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·30²/4 = 706.9 mm²
σ = P/A = 115·1000/706.9 = 162.7 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 162.7/70000 = 0.002324 = 2324 µε

Elongation

δ = ε·L = 0.002324·1000 mm = 2.324 mm

Answer

δ = 2.324 mm
2.Axial stress & strain

Answer: σ2 = 21.57 MPa

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Stiffnesses

k₁ = 200000·490.9/500 = 196350 N/mm
k₂ = 120000·1963/500 = 471239 N/mm

Share of segment 2

F2 = 60000·471239/667588 = 42353 N

Stress

σ2 = 42353/1963 = 21.57 MPa (compression)

Answer

σ2 = 21.57 MPa
3.Torsion of shafts

Answer: φ = 2.63 °

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(80⁴ − 40⁴)/32 = 3.77 × 10⁶ mm⁴

Angle of twist

T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 3 m = 3000 mm
G = 26 GPa = 26000 MPa
φ = TL/(GJ) = 1.5 × 10⁶·3000/(26000·3.77 × 10⁶) = 0.04591 rad
φ = 0.04591·180/π = 2.63°

Answer

φ = 2.63°
4.Torsion of shafts

Answer: φC = 2.026 °

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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 750 N·m
TAB = TC + TB = 750 + 400 = 1150 N·m

Twist of each segment

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 80 GPa = 80000 MPa
φAB = 1.15 × 10⁶·1500/(80000·1.272 × 10⁶) = 0.01695 rad
φBC = 750000·2500/(80000·1.272 × 10⁶) = 0.01842 rad

Twist of end C

φC = 0.01695 + 0.01842 = 0.03537 rad → |φC| = 2.026°

Answer

φC = 2.026°
5.Shear force & bending moment

Answer: C (x = 4 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 = 45 kN, R2 = 25 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 = 90, MC = 180 kN·m, MD = 150, ME = 100, MF = 50, MG = 0

Answer

Mark C: MC = 180 kN·m
6.Shear force & bending moment

Answer: Mmax = 7.385 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 = 5·5.5 = 27.5 kN, acting at x̄ = 2.75 m
ΣMA = 0: RB·4 = 5·5.5·2.75
RB = 75.63/4 = 18.91 kN
ΣFy = 0: RA = 27.5 − 18.91 = 8.594 kN

Zero shear

RA − w·x0 = 0 → x0 = 8.594/5 = 1.719 m

Maximum sagging moment

Mmax = RA·x0 − w·x0²/2 = RA²/(2w) = 8.594²/(2·5) = 7.385 kN·m

Answer

Mmax = 7.385 kN·m
7.Centroid & second moment of area

Answer: Ix = 66.1 × 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 two rectangles

Measure every y from the bottom edge.

flange: Af = 300·25 = 7500 mm², yf = 280 + 25/2 = 292.5 mm
web: Aw = 10·280 = 2800 mm², yw = 280/2 = 140 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (7500·292.5 + 2800·140)/10300 = 251 mm

Parallel-axis theorem

flange: 300·25³/12 + 7500·(292.5 − 251)² = 1.328 × 10⁷ mm⁴
web: 10·280³/12 + 2800·(140 − 251)² = 5.282 × 10⁷ mm⁴
Ix = 1.328 × 10⁷ + 5.282 × 10⁷ = 6.61 × 10⁷ mm⁴ = 66.1 × 10⁶ mm⁴

Answer

Ix = 66.1 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: Ix = 53.19 × 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 = 1900 mm², y = 185 mm
web: A = 2240 mm², y = 100 mm
bottom flange: A = 6000 mm², y = 10 mm

Centroid

ȳ = (1900·185 + 2240·100 + 6000·10)/10140 = 62.67 mm

Parallel-axis theorem

top flange: 190·10³/12 + 1900·122.3² = 2.845 × 10⁷ mm⁴
web: 14·160³/12 + 2240·37.33² = 7.9 × 10⁶ mm⁴
bottom flange: 300·20³/12 + 6000·52.67² = 1.685 × 10⁷ mm⁴
Ix = 2.845 × 10⁷ + 7.9 × 10⁶ + 1.685 × 10⁷
Ix = 5.319 × 10⁷ mm⁴ = 53.19 × 10⁶ mm⁴

Answer

Ix = 53.19 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τmax = 69.46 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 = √(((40 − −30)/2)² + 60²) = 69.46 MPa

Answer

τmax = 69.46 MPa
10.Mohr’s circle & stress transformation

Answer: τθ = 4.93 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

τθ = −−40·−0.6428 + 40·0.766 = 4.93 MPa

Answer

τθ = 4.93 MPa
11.Beam deflection

Answer: δmax = 26.35 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 = 2.53 × 10¹¹ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·2·4000⁴/(384·2.53 × 10¹¹) = 26.35 mm

Answer

δmax = 26.35 mm
12.Beam deflection

Answer: Ireq = 218.2 × 10⁶ mm⁴

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).

Deflection limit

δallow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·8·4000⁴/(384·11000·11.11) = 2.182 × 10⁸ mm⁴
Ireq = 218.2 × 10⁶ mm⁴

Answer

Ireq = 218.2 × 10⁶ mm⁴
13.Column buckling

Answer: Pcr = 334.2 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·533333/1050² = 334209 N = 334.2 kN

Answer

Pcr = 334.2 kN
14.Column buckling

Answer: d = 130.1 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 = 3·90000 = 270000 N
KL = 2·3000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·270000·6000²/(π³·70000))^(1/4) = 130.1 mm

Answer

d = 130.1 mm

Check: the slenderness KL/r is 184.5, so the column is slender and Euler applies.

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