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 = 100 kNd = 30 mmL = 3 mE = 117 GPa
δ = mm
2.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaP = 100 kN
σ2 = MPa
3.Torsion of shaftsMedium
A solid aluminium 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 = 750 N·md = 40 mmL = 0.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 100 kW at 1000 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
P = 100 kWN = 1000 rpmdi/do = 0.7τallow = 75 MPa
do = mm
5.Shear force & bending momentMedium
The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2 m from A. Find the bending moment Mx at the section x = 1 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.
L = 2 mP = 15 kN at 2 mw = 5 kN/m from 0 to 2 mx = 1 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.
20 kN @ 12 m10 kN @ 4 m8 kN/m, 0–10 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 120 mm × 200 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 = 120 mmh = 200 mmd = 30 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A 200 mm × 150 mm plate has a rectangular cut-out 30 mm wide and 30 mm high, centred on the vertical centre line with its centre 105 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 150 mmcut-out: 30 × 30 mmhole centre: y = 105 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −80 MPa and τₓᵧ = −40 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = −70 MPaσy = −80 MPaτxy = −40 MPaσz = 0
τabs = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 40 MPa, σᵧ = 20 MPa and τₓᵧ = 50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 40 MPaσy = 20 MPaτxy = 50 MPaθ = -45°
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 = 10 kN/mL = 4 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴
δmax = 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 = 30 kNw = 6 kN/mL = 3 mE = 200 GPa (steel)I = 84 × 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 = 2.5 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
14.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 60 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 60 kNn = 3L = 5 mE = 70 GPa (aluminium)
d = mm
3Solutions
1.Axial stress & strain
Answer: δ = 3.627 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·30²/4 = 706.9 mm²
σ = P/A = 100·1000/706.9 = 141.5 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 141.5/117000 = 0.001209 = 1209 µε
Elongation
δ = ε·L = 0.001209·3000 mm = 3.627 mm
Answer
δ = 3.627 mm
2.Axial stress & strain
Answer: σ2 = 54.68 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 70000·706.9/400 = 123700 N/mm
Share of segment 2
F2 = 100000·123700/320050 = 38650 N
Stress
σ2 = 38650/706.9 = 54.68 MPa (compression)
Answer
σ2 = 54.68 MPa
3.Torsion of shafts
Answer: φ = 3.288 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·500/(26000·251327) = 0.05739 rad
φ = 0.05739·180/π = 3.288°
Answer
φ = 3.288°
4.Torsion of shafts
Answer: do = 44.03 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 = 100000/104.7 = 954.9 N·m
Outer diameter from the allowable stress
T = 954.9 N·m = 954930 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·954930/(π·75·0.7599)) = 44.03 mm
di = 0.7·44.03 = 30.82 mm
Answer
do = 44.03 mm
A solid shaft would need d = 40.18 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.
Keep the free-end part
P: −15·1 = −15 kN·m
w: −5·1²/2 = −2.5 kN·m
Mx = −15 − 2.5 = −17.5 kN·m
Answer
Mx = −17.5 kN·m
Negative means hogging: tension on the top face.
6.Shear force & bending moment
Answer: C (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 = 42 kN, R2 = 68 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 = 68, MC = 104 kN·m, MD = 88, ME = 40, MF = −40, MG = 0
Answer
Mark C: MC = 104 kN·m
7.Centroid & second moment of area
Answer: Ix = 78.79 × 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 = (24000·100 − 706.9·60)/(24000 − 706.9) = 101.2 mm
Full plate about ȳ
120·200³/12 + 24000·−1.214² = 80.04 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·−41.21² = 1.24 × 10⁶ mm⁴
Subtract
Ix = 78.79 × 10⁶ mm⁴
Answer
Ix = 78.79 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 55.35 × 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 − 900·105)/(30000 − 900) = 74.07 mm
Full plate about ȳ
200·150³/12 + 30000·0.9278² = 56.28 × 10⁶ mm⁴
Hole about ȳ
30·30³/12 + 900·30.93² = 0.9284 × 10⁶ mm⁴
Subtract
Ix = 55.35 × 10⁶ mm⁴
Answer
Ix = 55.35 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: τabs = 57.66 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = −75 MPa, R = √(5² + (−40)²) = 40.31 MPa
σ1 = −34.69 MPa, σ2 = −115.3 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(80.62, 34.69, 115.3)/2 = 57.66 MPa
Answer
τabs = 57.66 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
10.Mohr’s circle & stress transformation
Answer: τθ = 10 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 = −10·−1 + 50·0 = 10 MPa
Answer
τθ = 10 MPa
11.Beam deflection
Answer: δmax = 5.669 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 = 10 N/mm, L = 4000 mm
EI = 5.88 × 10¹² N·mm²
Deflection
δmax = 5wL⁴/(384EI) = 5·10·4000⁴/(384·5.88 × 10¹²) = 5.669 mm
Answer
δmax = 5.669 mm
12.Beam deflection
Answer: δmax = 1.381 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 = 30000 N, w = 6 N/mm, L = 3000 mm, EI = 1.68 × 10¹³ N·mm²
Each load on its own
δP = PL³/(48EI) = 30000·3000³/(48·1.68 × 10¹³) = 1.004 mm
δw = 5wL⁴/(384EI) = 5·6·3000⁴/(384·1.68 × 10¹³) = 0.3767 mm
Add them
δmax = 1.004 + 0.3767 = 1.381 mm
Answer
δmax = 1.381 mm
13.Column buckling
Answer: Pcr = 12.44 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 = 50·30³/12 = 112500 mm⁴
Effective length
KL = 1·2500 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·112500/2500² = 12436 N = 12.44 kN
Answer
Pcr = 12.44 kN
14.Column buckling
Answer: d = 107.3 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 = 3·60000 = 180000 N
KL = 1·5000 = 5000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·180000·5000²/(π³·70000))^(1/4) = 107.3 mm
Answer
d = 107.3 mm
Check: the slenderness KL/r is 186.3, so the column is slender and Euler applies.
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