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 elongation δ of the copper bar under the axial load.
P = 40 kNd = 16 mmL = 1.5 mE = 117 GPa
δ = mm
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 = 60 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, brass (E = 100 GPa)
σmax = MPa
3.Torsion of shaftsMedium
A hollow brass shaft (outer diameter 50 mm, inner diameter 35 mm, length 1.5 m) is fixed at one end and twisted by a torque of 1000 N·m at the other. Find the angle of twist φ in degrees.
T = 1000 N·mdo = 50 mmdi = 35 mmL = 1.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 3000 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 3000 rpmdo = 75 mmdi = 45 mm
τmax = MPa
5.Shear force & bending momentMedium
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 @ 1 m35 kN @ 5 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)
6.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
L = 6 mP = 40 kN at 1 mw = 12 kN/m from 0 to 6 m
Mmax = kN·m
7.Centroid & second moment of areaMedium
A 120 mm × 300 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 180 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 300 mmd = 60 mmhole centre: y = 180 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A 150 mm × 300 mm plate has a rectangular cut-out 40 mm wide and 20 mm high, centred on the vertical centre line with its centre 210 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 300 mmcut-out: 40 × 20 mmhole centre: y = 210 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 80 MPa, σᵧ = 70 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 80 MPaσy = 70 MPaτxy = −50 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −60 MPa, σᵧ = −80 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.
σx = −60 MPaσy = −80 MPaτxy = 40 MPaθ = 25°
σθ = MPa
11.Beam deflectionMedium
A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.
w = 2 kN/mL = 4 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴
δ = mm
12.Beam deflectionHard
A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction RB at the prop.
L = 5 mw = 5 kN/m
RB = kN
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 = 5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
14.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 80 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 80 kNn = 2.5L = 5 mE = 200 GPa (steel)
d = mm
3Solutions
1.Axial stress & strain
Answer: δ = 2.551 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·16²/4 = 201.1 mm²
σ = P/A = 40·1000/201.1 = 198.9 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 198.9/117000 = 0.0017 = 1700 µε
Elongation
δ = ε·L = 0.0017·1500 mm = 2.551 mm
Answer
δ = 2.551 mm
2.Axial stress & strain
Answer: σmax = 191 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·40²/4 = 1257 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 47.75 MPa
σ2 = 191 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 191 MPa
Answer
σmax = 191 MPa
3.Torsion of shafts
Answer: φ = 4.726 °
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 − 35⁴)/32 = 466269 mm⁴
Angle of twist
T = 1000 N·m = 1 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 1 × 10⁶·1500/(39000·466269) = 0.08249 rad
φ = 0.08249·180/π = 4.726°
Answer
φ = 4.726°
4.Torsion of shafts
Answer: τmax = 4.415 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·3000/60 = 314.2 rad/s
T = P/ω = 100000/314.2 = 318.3 N·m
Polar second moment of area
J = π(75⁴ − 45⁴)/32 = 2.704 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 318310·37.5/2.704 × 10⁶ = 4.415 MPa
Answer
τmax = 4.415 MPa
5.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 = 22.5 kN, R2 = 32.5 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 = 22.5, MC = 25, MD = 27.5, ME = 30, MF = 32.5 kN·m, MG = 0
Answer
Mark F: MF = 32.5 kN·m
6.Shear force & bending moment
Answer: Mmax = 75.85 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.
UDL resultant: W = w·ℓ = 12·6 = 72 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 40·1 + 12·6·3
RB = 256/6 = 42.67 kN
ΣFy = 0: RA = 112 − 42.67 = 69.33 kN
Where is V = 0?
RA − P − w·x0 = 0 → x0 = (69.33 − 40)/12 = 2.444 m
Maximum moment
Mmax = RA·x0 − P·(x0 − 1) − w·x0²/2 = 75.85 kN·m
Answer
Mmax = 75.85 kN·m
It occurs at x = 2.444 m from A.
7.Centroid & second moment of area
Answer: Ix = 266.6 × 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 − 2827·180)/(36000 − 2827) = 147.4 mm
Full plate about ȳ
120·300³/12 + 36000·2.557² = 270.2 × 10⁶ mm⁴
Hole about ȳ
π·60⁴/64 + 2827·32.56² = 3.633 × 10⁶ mm⁴
Subtract
Ix = 266.6 × 10⁶ mm⁴
Answer
Ix = 266.6 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 334.5 × 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 = (45000·150 − 800·210)/(45000 − 800) = 148.9 mm
Full plate about ȳ
150·300³/12 + 45000·1.086² = 337.6 × 10⁶ mm⁴
Hole about ȳ
40·20³/12 + 800·61.09² = 3.012 × 10⁶ mm⁴
Subtract
Ix = 334.5 × 10⁶ mm⁴
Answer
Ix = 334.5 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: τmax = 50.25 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 = √(((80 − 70)/2)² + −50²) = 50.25 MPa
Answer
τmax = 50.25 MPa
10.Mohr’s circle & stress transformation
Answer: σθ = −32.93 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
σθ = −70 + 10·0.6428 + 40·0.766 = −32.93 MPa
Answer
σθ = −32.93 MPa
11.Beam deflection
Answer: δ = 5.614 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 = 1.14 × 10¹³ N·mm²
Deflection
δ = wL⁴/(8EI) = 2·4000⁴/(8·1.14 × 10¹³) = 5.614 mm
Answer
δ = 5.614 mm
12.Beam deflection
Answer: RB = 9.375 kN
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). For a statically indeterminate beam, remove a redundant support and require the deflection there to be zero (compatibility).
Remove the prop
Without the prop the beam is a cantilever. Its free end would deflect under the load.
δload = wL⁴/(8EI)
Compatibility at the prop
The prop force RB alone lifts the free end by RBL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.
RB·L³/(3EI) = wL⁴/(8EI) → RB = 3wL/8 = 3·5·5/8 = 9.375 kN
Answer
RB = 9.375 kN
13.Column buckling
Answer: Pcr = 17.58 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/5000² = 17581 N = 17.58 kN
Answer
Pcr = 17.58 kN
14.Column buckling
Answer: d = 70.91 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 = 2.5·80000 = 200000 N
KL = 0.7·5000 = 3500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·3500²/(π³·200000))^(1/4) = 70.91 mm
Answer
d = 70.91 mm
Check: the slenderness KL/r is 197.4, so the column is slender and Euler applies.
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