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 copper bar, in microstrain (µε = 10⁻⁶).
P = 10 kNd = 30 mmL = 0.5 mE = 117 GPa
ε = µε
2.Axial stress & strainHard
A steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 300 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 30 kN
σ2 = MPa
3.Torsion of shaftsMedium
A stepped solid steel shaft is fixed at A. Segment AB has diameter 60 mm and length 1.2 m; segment BC has diameter 40 mm and length 1 m. A torque of 750 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 750 N·md1 = 60 mm, L1 = 1.2 md2 = 40 mm, L2 = 1 m
τmax = MPa
4.Torsion of shaftsHard
A solid steel shaft transmits 5 kW at 1000 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.
P = 5 kWN = 1000 rpmτallow = 80 MPaθallow = 0.5°/mG = 80 GPa
d = mm
5.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mw = 3 kN/m from 0 to 8 mx = 2 m
Mx = 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.
60 kN @ 12 m25 kN @ 2 m12 kN/m, 0–8 m
- 1A (x = 0 m)
- 2B (x = 2 m)
- 3C (x = 4 m)
- 4D (x = 6 m)
- 5E (x = 8 m)
- 6F (x = 10 m)
- 7G (x = 12 m)
7.Centroid & second moment of areaMedium
A 150 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 97.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 150 mmd = 30 mmhole centre: y = 97.5 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A 100 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 144 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 240 mmcut-out: 50 × 40 mmhole centre: y = 144 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 90 MPa, σᵧ = 10 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 90 MPaσy = 10 MPaτxy = 10 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = 100 MPa and τₓᵧ = −60 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −20 MPaσy = 100 MPaτxy = −60 MPaσY = 250 MPa
n =
11.Beam deflectionMedium
A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.
w = 15 kN/mL = 4 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴
θA = rad
12.Beam deflectionHard
A simply supported beam carries a point load P at b = 1 m from support B. Find the maximum deflection δmax of the beam.
P = 8 kNb = 1 mL = 8 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
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 mE = 200 GPa (steel)b × h = 25 × 80 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 = 20 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: ε = 120.9 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 = 10·1000/706.9 = 14.15 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 14.15/117000 = 0.0001209 = 120.9 µε
Answer
ε = 120.9 µε
2.Axial stress & strain
Answer: σ2 = 10.23 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 = 200000·1257/300 = 837758 N/mm
k₂ = 100000·1257/200 = 628319 N/mm
Share of segment 2
F2 = 30000·628319/1.466 × 10⁶ = 12857 N
Stress
σ2 = 12857/1257 = 10.23 MPa (compression)
Answer
σ2 = 10.23 MPa
3.Torsion of shafts
Answer: τmax = 59.68 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.
Stress in each segment
T = 750 N·m = 750000 N·mm
τAB = 16·750000/(π·60³) = 17.68 MPa
τBC = 16·750000/(π·40³) = 59.68 MPa
Answer
τmax = 59.68 MPa
4.Torsion of shafts
Answer: d = 28.89 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·1000/60 = 104.7 rad/s
T = P/ω = 5000/104.7 = 47.75 N·m
Diameter for strength
T = 47.75 N·m = 47746 N·mm
dτ = ∛(16·47746/(π·80)) = 14.49 mm
Diameter for stiffness
θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·47746/(π·80000·8.727 × 10⁻⁶))1/4 = 28.89 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(14.49, 28.89) = 28.89 mm
Answer
d = 28.89 mm
Here stiffness governs.
5.Shear force & bending moment
Answer: Mx = 18 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 = 3·8 = 24 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 3·8·4
RB = 96/8 = 12 kN
ΣFy = 0: RA = 24 − 12 = 12 kN
Cut at x = 2 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 = 12·2 − 3·2·1 = 18 kN·m
Answer
Mx = 18 kN·m
6.Shear force & bending moment
Answer: E (x = 8 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 = 36.75 kN, R2 = 144.3 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 = 49.5, MC = 1, MD = −95.5, ME = −240 kN·m, MF = −120, MG = 0
Answer
Mark E: ME = −240 kN·m
7.Centroid & second moment of area
Answer: Ix = 41.78 × 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 = (22500·75 − 706.9·97.5)/(22500 − 706.9) = 74.27 mm
Full plate about ȳ
150·150³/12 + 22500·0.7298² = 42.2 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·23.23² = 0.4212 × 10⁶ mm⁴
Subtract
Ix = 41.78 × 10⁶ mm⁴
Answer
Ix = 41.78 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 113.7 × 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 = (24000·120 − 2000·144)/(24000 − 2000) = 117.8 mm
Full plate about ȳ
100·240³/12 + 24000·2.182² = 115.3 × 10⁶ mm⁴
Hole about ȳ
50·40³/12 + 2000·26.18² = 1.638 × 10⁶ mm⁴
Subtract
Ix = 113.7 × 10⁶ mm⁴
Answer
Ix = 113.7 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: τmax = 41.23 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 = √(((90 − 10)/2)² + 10²) = 41.23 MPa
Answer
τmax = 41.23 MPa
10.Mohr’s circle & stress transformation
Answer: n = 1.473
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 40 MPa, R = 84.85 MPa
σ1 = 124.9 MPa, σ2 = −44.85 MPa, σ3 = 0
Tresca
σmax − σmin = 169.7 MPa
n = 250/169.7 = 1.473
Answer
n = 1.473
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
11.Beam deflection
Answer: θA = 0.02273 rad
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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.
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 = 15 N/mm, L = 4000 mm
EI = 1.76 × 10¹² N·mm²
End slope
θA = wL³/(24EI) = 15·4000³/(24·1.76 × 10¹²) = 0.02273 rad
Answer
θA = 0.02273 rad
That is 1.302°.
12.Beam deflection
Answer: δmax = 24.3 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.
P = 8000 N, b = 1000 mm, L = 8000 mm
EI = 1.32 × 10¹² N·mm²
Position of the maximum
Setting the slope in the longer segment to zero gives the position.
x = √((L² − b²)/3) = √((8000² − 1000²)/3) = 4583 mm
Maximum deflection
δmax = Pb(L² − b²)3/2/(9√3·EIL) = 24.3 mm
Answer
δmax = 24.3 mm
13.Column buckling
Answer: Pcr = 205.6 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 = 80·25³/12 = 104167 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·104167/1000² = 205617 N = 205.6 kN
Answer
Pcr = 205.6 kN
14.Column buckling
Answer: σcr = 16.11 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·20⁴/64 = 7854 mm⁴
Radius of gyration and slenderness
A = 314.2 mm²
r = √(I/A) = √(7854/314.2) = 5 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/5 = 350
Critical stress
σcr = π²E/(KL/r)² = π²·200000/350² = 16.11 MPa
Answer
σcr = 16.11 MPa
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