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
Mechanics of Materials: cram pack
Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
1Formulas and methods
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
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- 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
- Find the support reactions from equilibrium of the whole beam.
- Cut the beam at the section you need and keep one side.
- Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
- 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
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- 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
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 20 kNd = 30 mmL = 1.5 mE = 70 GPa
ε = µε
2.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 120 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
3.Torsion of shaftsMedium
A solid brass 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 = 400 N·md = 50 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 0.8 m from A and 3 m from B. Find the angle of rotation φC of section C, in degrees.
d = 40 mmT0 = 250 N·ma = 0.8 mb = 3 mG = 39 GPa
φC = °
5.Shear force & bending momentMedium
The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.
L = 6 mP1 = 40 kN at 1 mP2 = 65 kN at 4 m
Mmax = kN·m
6.Shear force & bending momentHard
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 @ 0 m35 kN @ 6 m
- 1A (x = 0 m)
- 2B (x = 1 m)
- 3C (x = 2 m)
- 4D (x = 3 m)
- 5E (x = 4 m)
- 6F (x = 5 m)
- 7G (x = 6 m)
7.Centroid & second moment of areaMedium
A channel section lies with its open side up: a base plate 220 mm wide with two upright legs, 130 mm high overall. Base and legs are all 16 mm thick. Find its second moment of area Ix about the horizontal axis x through its centroid.
b = 220 mmh = 130 mmt = 16 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A welded I-section has a top flange 120 × 22 mm, a web 390 mm high and 8 mm thick, and a bottom flange 180 × 14 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.
top flange: 120 × 22 mmweb: 8 × 390 mmbottom flange: 180 × 14 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 70 MPa, σᵧ = −20 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 70 MPaσy = −20 MPaτxy = 40 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −20 MPa, σᵧ = 90 MPa and τₓᵧ = 50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −20 MPaσy = 90 MPaτxy = 50 MPaθ = -30°
τθ = MPa
11.Beam deflectionMedium
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 = 5 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
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 = 18 kNL = 6 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 25 mm
Pcr = kN
14.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: ε = 404.2 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·30²/4 = 706.9 mm²
σ = P/A = 20·1000/706.9 = 28.29 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 28.29/70000 = 0.0004042 = 404.2 µε
Answer
ε = 404.2 µε
2.Axial stress & strain
Answer: σmax = 382 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·20²/4 = 314.2 mm²
σ1 = 124.7 MPa
σ2 = 382 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 382 MPa
Answer
σmax = 382 MPa
3.Torsion of shafts
Answer: φ = 0.9577 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·1000/(39000·613592) = 0.01672 rad
φ = 0.01672·180/π = 0.9577°
Answer
φ = 0.9577°
4.Torsion of shafts
Answer: φC = 0.923 °
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 = 250
TA·a = TB·b
TA = 250·3/3.8 = 197.4 N·m
TB = 250·0.8/3.8 = 52.63 N·m
Rotation of C
Use segment AC: it is fixed at A and carries TA.
J = π·40⁴/32 = 251327 mm⁴
G = 39 GPa = 39000 MPa
φC = TA·a/(GJ) = 197368·800/(39000·251327) = 0.01611 rad = 0.923°
Answer
φC = 0.923°
5.Shear force & bending moment
Answer: Mmax = 100 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.
ΣMA = 0: RB·6 = 40·1 + 65·4
RB = 300/6 = 50 kN
ΣFy = 0: RA = 105 − 50 = 55 kN
Moment under each load
Keep the part left of each load.
under P1: M = RA·1 = 55·1 = 55 kN·m
under P2: M = RB·2 = 50·2 = 100 kN·m
Largest value
Mmax = 100 kN·m (under P2)
Answer
Mmax = 100 kN·m
6.Shear force & bending moment
Answer: E (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 = 3.333 kN, R2 = 51.67 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 = −20, MC = −36.67, MD = −53.33, ME = −70 kN·m, MF = −35, MG = 0
Answer
Mark E: ME = −70 kN·m
7.Centroid & second moment of area
Answer: Ix = 11.59 × 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 rectangles
Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.
base: A1 = 220·16 = 3520 mm², y1 = 8 mm
two legs: A2 = 2·16·114 = 3648 mm², y2 = 16 + 114/2 = 73 mm
Centroid
ȳ = (3520·8 + 3648·73)/7168 = 41.08 mm
Parallel-axis theorem
base: 220·16³/12 + 3520·(8 − 41.08)² = 3.927 × 10⁶ mm⁴
two legs: 2·16·114³/12 + 3648·(73 − 41.08)² = 7.668 × 10⁶ mm⁴
Ix = 3.927 × 10⁶ + 7.668 × 10⁶
Ix = 1.159 × 10⁷ mm⁴ = 11.59 × 10⁶ mm⁴
Answer
Ix = 11.59 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 254.4 × 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 = 2640 mm², y = 415 mm
web: A = 3120 mm², y = 209 mm
bottom flange: A = 2520 mm², y = 7 mm
Centroid
ȳ = (2640·415 + 3120·209 + 2520·7)/8280 = 213.2 mm
Parallel-axis theorem
top flange: 120·22³/12 + 2640·201.8² = 1.076 × 10⁸ mm⁴
web: 8·390³/12 + 3120·4.203² = 3.96 × 10⁷ mm⁴
bottom flange: 180·14³/12 + 2520·206.2² = 1.072 × 10⁸ mm⁴
Ix = 1.076 × 10⁸ + 3.96 × 10⁷ + 1.072 × 10⁸
Ix = 2.544 × 10⁸ mm⁴ = 254.4 × 10⁶ mm⁴
Answer
Ix = 254.4 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: τmax = 60.21 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.
Radius of Mohr’s circle
τmax = R = √(((70 − −20)/2)² + 40²) = 60.21 MPa
Answer
τmax = 60.21 MPa
10.Mohr’s circle & stress transformation
Answer: τθ = −22.63 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 = −−55·−0.866 + 50·0.5 = −22.63 MPa
Answer
τθ = −22.63 MPa
11.Beam deflection
Answer: δmax = 1.453 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
EI = 1.12 × 10¹³ N·mm²
Deflection
δmax = 5wL⁴/(384EI) = 5·2·5000⁴/(384·1.12 × 10¹³) = 1.453 mm
Answer
δmax = 1.453 mm
12.Beam deflection
Answer: δmax = 30 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 = 18000 N, L = 6000 mm
EI = 4.6 × 10¹² N·mm²
One load at L/3
δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 2000 mm
δ₁ = 15 mm
Add both loads
δmax = 2·δ₁ = 23PL³/(648EI) = 30 mm
Answer
δmax = 30 mm
13.Column buckling
Answer: Pcr = 3.312 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·19175/2000² = 3312 N = 3.312 kN
Answer
Pcr = 3.312 kN
14.Column buckling
Answer: σcr = 36.26 MPa
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
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/7.5 = 233.3
Critical stress
σcr = π²E/(KL/r)² = π²·200000/233.3² = 36.26 MPa
Answer
σcr = 36.26 MPa
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