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
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa). Both are 800 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 800 mm
δ = mm
2.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 120 MPatube: Ø80/Ø60 mm, E = 110 GPa, σallow = 50 MPa
Pmax = kN
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 40 N·m. The two ends of a gauge length of 300 mm rotate 4.22° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 40 N·md = 12 mmL = 300 mmφ = 4.22°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1000 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1000 rpmdo = 75 mmdi = 60 mm
τmax = MPa
5.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mw = 10 kN/m from 0 to 8 mx = 0.5 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 @ 0 m45 kN @ 12 m5 kN/m, 2–10 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 100 mm × 200 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 200 mmd = 50 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A symmetric I-section has flanges 190 mm wide and 14 mm thick, an overall depth of 420 mm and a web 8 mm thick. It carries a bending moment of 30 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
B = 190 mmH = 420 mmtf = 14 mmtw = 8 mmM = 30 kN·m
σj = MPa
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −60 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −60 MPaσy = −70 MPaτxy = −30 MPa
θp = °
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 10 MPa, σᵧ = 0 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 10 MPaσy = 0 MPaτxy = −30 MPaθ = 50°
σθ = MPa
11.Beam deflectionMedium
A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δP under the load.
P = 23 kNa = 1 mL = 8 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴
δP = mm
12.Beam deflectionHard
A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δmax.
P = 25 kNw = 6 kN/mL = 6 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)b × h = 40 × 50 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 100 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 100 kNn = 2L = 3 mE = 200 GPa (steel)
d = mm
3Solutions
1.Axial stress & strain
Answer: δ = 0.1787 mm
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.
Areas
Ar = π·30²/4 = 706.9 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 100·10³·3063 = 4.477 × 10⁸ N
Shortening
δ = 100000·800/4.477 × 10⁸ = 0.1787 mm
Answer
δ = 0.1787 mm
2.Axial stress & strain
Answer: Pmax = 224.2 kN
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.
Areas
Ar = π·40²/4 = 1257 mm²
At = π(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 110·10³·2199 = 4.932 × 10⁸ N
Governing strain
ε = min(0.0006, 0.0004545) = 0.0004545 (bronze)
Largest load
Pmax = 0.0004545·4.932 × 10⁸ = 224195 N = 224.2 kN
Answer
Pmax = 224.2 kN
3.Torsion of shafts
Answer: G = 80.03 GPa
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 moment and angle
J = π·12⁴/32 = 2036 mm⁴
φ = 4.22·π/180 = 0.07365 rad
Shear modulus
T = 40 N·m = 40000 N·mm
G = TL/(Jφ) = 40000·300/(2036·0.07365) = 80033 MPa
G = 80.03 GPa
Answer
G = 80.03 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 19.53 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·1000/60 = 104.7 rad/s
T = P/ω = 100000/104.7 = 954.9 N·m
Polar second moment of area
J = π(75⁴ − 60⁴)/32 = 1.834 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·37.5/1.834 × 10⁶ = 19.53 MPa
Answer
τmax = 19.53 MPa
5.Shear force & bending moment
Answer: Mx = 18.75 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 = 10·8 = 80 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 10·8·4
RB = 320/8 = 40 kN
ΣFy = 0: RA = 80 − 40 = 40 kN
Cut at x = 0.5 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 = 40·0.5 − 10·0.5·0.25 = 18.75 kN·m
Answer
Mx = 18.75 kN·m
6.Shear force & bending moment
Answer: B (x = 2 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 = 83.75 kN, R2 = 61.25 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 = −120 kN·m, MC = −82.5, MD = −65, ME = −67.5, MF = −90, MG = 0
Answer
Mark B: MB = −120 kN·m
7.Centroid & second moment of area
Answer: Ix = 62.88 × 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 = (20000·100 − 1963·60)/(20000 − 1963) = 104.4 mm
Full plate about ȳ
100·200³/12 + 20000·−4.354² = 67.05 × 10⁶ mm⁴
Hole about ȳ
π·50⁴/64 + 1963·−44.35² = 4.17 × 10⁶ mm⁴
Subtract
Ix = 62.88 × 10⁶ mm⁴
Answer
Ix = 62.88 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σj = 22.66 MPa
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.
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 = 420 − 2·14 = 392 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (190·420³ − 182·392³)/12
Ix = 2.595 × 10⁸ mm⁴
Bending stress
y = 210 − 14 = 196 mm
σ = M·y/I = 3 × 10⁷·196/2.595 × 10⁸ = 22.66 MPa
Answer
σj = 22.66 MPa
9.Mohr’s circle & stress transformation
Answer: θp = −40.27 °
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.
Angle on the circle
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·−30, −60 − −70) = −80.54°
Angle on the element
θp = −40.27°
Answer
θp = −40.27°
Positive angles are counter-clockwise.
10.Mohr’s circle & stress transformation
Answer: σθ = −25.41 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 = 5 + 5·−0.1736 + −30·0.9848 = −25.41 MPa
Answer
σθ = −25.41 MPa
11.Beam deflection
Answer: δP = 2.795 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 = 23000 N, a = 1000 mm, b = 7000 mm, L = 8000 mm
EI = 1.68 × 10¹³ N·mm²
Deflection under the load
δP = Pa²b²/(3EIL) = 23000·1000²·7000²/(3·1.68 × 10¹³·8000) = 2.795 mm
Answer
δP = 2.795 mm
12.Beam deflection
Answer: δmax = 8.906 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 = 25000 N, w = 6 N/mm, L = 6000 mm, EI = 2.4 × 10¹³ N·mm²
Each load on its own
δP = PL³/(48EI) = 25000·6000³/(48·2.4 × 10¹³) = 4.688 mm
δw = 5wL⁴/(384EI) = 5·6·6000⁴/(384·2.4 × 10¹³) = 4.219 mm
Add them
δmax = 4.688 + 4.219 = 8.906 mm
Answer
δmax = 8.906 mm
13.Column buckling
Answer: Pcr = 11.51 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 = 50·40³/12 = 266667 mm⁴
Effective length
KL = 2·2000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·266667/4000² = 11515 N = 11.51 kN
Answer
Pcr = 11.51 kN
14.Column buckling
Answer: d = 54.93 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 = 2·100000 = 200000 N
KL = 0.7·3000 = 2100 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·2100²/(π³·200000))^(1/4) = 54.93 mm
Answer
d = 54.93 mm
Check: the slenderness KL/r is 152.9, so the column is slender and Euler applies.
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