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 bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 500 mm, E = 70 GPaδ = 0.25 mm
P = kN
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 = 10 kNsegment 1: Ø40 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 GPa)
σmax = MPa
3.Torsion of shaftsMedium
A solid steel 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 = 200 N·md = 50 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 15 kW at 300 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.
P = 15 kWN = 300 rpmdi/do = 0.6τallow = 50 MPa
do = mm
5.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mw = 5 kN/m from 0 to 8 mx = 7 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.
30 kN @ 0 m25 kN @ 12 m40 kN @ 6 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 200 mm × 150 mm plate has a circular hole of diameter 30 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 = 200 mmh = 150 mmd = 30 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A 200 mm × 240 mm plate has a rectangular cut-out 80 mm wide and 50 mm high, centred on the vertical centre line with its centre 96 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 240 mmcut-out: 80 × 50 mmhole centre: y = 96 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point has σₓ = −50 MPa, σᵧ = 20 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −50 MPaσy = 20 MPaτxy = −50 MPa
θs = °
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 120 MPa, σᵧ = 80 MPa and τₓᵧ = 30 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 = 120 MPaσy = 80 MPaτxy = 30 MPaθ = -20°
τθ = MPa
11.Beam deflectionMedium
A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.
P = 2 kNw = 12 kN/mL = 3 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴
δ = mm
12.Beam deflectionHard
A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth hreq of the section.
w = 3 kN/mL = 3 mE = 11 GPa (timber)b = 125 mm
hreq = mm
13.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 200 GPa (steel)b × h = 25 × 60 mm
Pcr = kN
14.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 70 GPa (aluminium)d = 25 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: P = 24.99 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
A₁ = 314.2 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·1963) = 3.638 × 10⁻⁶ mm/N
Load
P = 0.25/1 × 10⁻⁵ = 24990 N = 24.99 kN
Answer
P = 24.99 kN
2.Axial stress & strain
Answer: σmax = 88.42 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·12²/4 = 113.1 mm²
σ1 = 7.958 MPa
σ2 = 88.42 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 88.42 MPa
Answer
σmax = 88.42 MPa
3.Torsion of shafts
Answer: φ = 0.3502 °
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) = 200·1000·1500/(80000·613592) = 0.006112 rad
φ = 0.006112·180/π = 0.3502°
Answer
φ = 0.3502°
4.Torsion of shafts
Answer: do = 38.23 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·300/60 = 31.42 rad/s
T = P/ω = 15000/31.42 = 477.5 N·m
Outer diameter from the allowable stress
T = 477.5 N·m = 477465 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·477465/(π·50·0.8704)) = 38.23 mm
di = 0.6·38.23 = 22.94 mm
Answer
do = 38.23 mm
A solid shaft would need d = 36.5 mm; the hollow one is a little larger but much lighter.
5.Shear force & bending moment
Answer: Mx = 17.5 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 = 5·8 = 40 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 5·8·4
RB = 160/8 = 20 kN
ΣFy = 0: RA = 40 − 20 = 20 kN
Cut at x = 7 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 = 20·7 − 5·7·3.5 = 17.5 kN·m
Answer
Mx = 17.5 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 = 51.25 kN, R2 = 43.75 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 = −60 kN·m, MC = −17.5, MD = 25, ME = −12.5, MF = −50, MG = 0
Answer
Mark B: MB = −60 kN·m
7.Centroid & second moment of area
Answer: Ix = 56.05 × 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 = (30000·75 − 706.9·60)/(30000 − 706.9) = 75.36 mm
Full plate about ȳ
200·150³/12 + 30000·−0.362² = 56.25 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·−15.36² = 0.2066 × 10⁶ mm⁴
Subtract
Ix = 56.05 × 10⁶ mm⁴
Answer
Ix = 56.05 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 227.1 × 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 = (48000·120 − 4000·96)/(48000 − 4000) = 122.2 mm
Full plate about ȳ
200·240³/12 + 48000·−2.182² = 230.6 × 10⁶ mm⁴
Hole about ȳ
80·50³/12 + 4000·−26.18² = 3.575 × 10⁶ mm⁴
Subtract
Ix = 227.1 × 10⁶ mm⁴
Answer
Ix = 227.1 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: θs = 72.5 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −50/61.03 = −0.8192, sin 2θs = −(−35)/61.03 = 0.5735
2θs = atan2(35, −50) = 145°
Angle on the element
θs = 72.5°
check: θs − θp = 72.5 − (−62.5) = 135°
Answer
θs = 72.5°
Positive angles are counter-clockwise.
10.Mohr’s circle & stress transformation
Answer: τθ = 35.84 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 = −20·−0.6428 + 30·0.766 = 35.84 MPa
Answer
τθ = 35.84 MPa
11.Beam deflection
Answer: δ = 8.304 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 = 2000 N, w = 12 N/mm, L = 3000 mm
EI = 1.68 × 10¹³ N·mm²
Each load on its own
δP = PL³/(3EI) = 2000·3000³/(3·1.68 × 10¹³) = 1.071 mm
δw = wL⁴/(8EI) = 12·3000⁴/(8·1.68 × 10¹³) = 7.232 mm
Add them
δ = 1.071 + 7.232 = 8.304 mm
Answer
δ = 8.304 mm
12.Beam deflection
Answer: hreq = 132 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 = 3 N/mm, L = 3000 mm, E = 11000 N/mm²
Deflection limit
δallow = L/250 = 3000/250 = 12 mm
Required second moment of area
Ireq = 5wL⁴/(384·E·δallow) = 5·3·3000⁴/(384·11000·12) = 2.397 × 10⁷ mm⁴
Required depth
bh³/12 = Ireq → h = (12·2.397 × 10⁷/125)1/3 = 132 mm
Answer
hreq = 132 mm
In practice, choose the next standard depth above this value.
13.Column buckling
Answer: Pcr = 24.67 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 = 60·25³/12 = 78125 mm⁴
Effective length
KL = 1·2500 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·78125/2500² = 24674 N = 24.67 kN
Answer
Pcr = 24.67 kN
14.Column buckling
Answer: σcr = 6.747 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/6.25 = 320
Critical stress
σcr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa
Answer
σcr = 6.747 MPa
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