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 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 60 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.
P = 300 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 100 GPaL = 300 mm
δ = mm
2.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 500 mm, E = 110 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: bronze, Ø30 mm, 500 mm, E = 110 GPaP = 100 kN
σ2 = MPa
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.
T = 750 N·mdi/do = 0.5τallow = 50 MPa
do = mm
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 2.5 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 200 N·mTC = 750 N·mAB = 2.5 mBC = 0.8 m
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.
50 kN @ 8 m10 kN/m, 4–12 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)
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.
45 kN @ 0 m30 kN @ 6 m15 kN @ 2 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)
7.Centroid & second moment of areaMedium
A 150 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 = 150 mmh = 200 mmd = 30 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 40 kN·m. Find the bending stress σbottom at the bottom edge.
bf = 100 mmtf = 20 mmtw = 18 mmhw = 300 mmM = 40 kN·m
σbottom = MPa
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 110 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 110 MPaσy = −20 MPaτxy = 30 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = 100 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.
σx = 70 MPaσy = 100 MPaτxy = 10 MPaθ = -20°
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 = 8 kN/mL = 5 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 point load P at midspan. EI is constant. Find the reaction RB at the prop.
L = 3 mP = 65 kN at 1.5 m
RB = kN
13.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
14.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 95 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 95 kNn = 3L = 3 mE = 70 GPa (aluminium)
d = mm
3Solutions
1.Axial stress & strain
Answer: δ = 0.1592 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·20²/4 = 314.2 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 100·10³·5027 = 5.655 × 10⁸ N
Shortening
δ = 300000·300/5.655 × 10⁸ = 0.1592 mm
Answer
δ = 0.1592 mm
2.Axial stress & strain
Answer: σ2 = 62.53 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 = 110000·706.9/500 = 155509 N/mm
Share of segment 2
F2 = 100000·155509/351858 = 44196 N
Stress
σ2 = 44196/706.9 = 62.53 MPa (compression)
Answer
σ2 = 62.53 MPa
3.Torsion of shafts
Answer: do = 43.35 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.
Outer diameter from the allowable stress
T = 750 N·m = 750000 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·750000/(π·50·0.9375)) = 43.35 mm
di = 0.5·43.35 = 21.68 mm
Answer
do = 43.35 mm
A solid shaft would need d = 42.43 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 38.71 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC + TB = 750 + 200 = 950 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 950 N·m
τmax = 16·950000/(π·50³) = 38.71 MPa
Answer
τmax = 38.71 MPa
5.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 = 43.33 kN, R2 = 86.67 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 = 86.67, MC = 173.3, MD = 240, ME = 266.7 kN·m, MF = 153.3, MG = 0
Answer
Mark E: ME = 266.7 kN·m
6.Shear force & bending moment
Answer: E (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 = 50 kN, R2 = 40 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 = −45, MC = −40, MD = −50, ME = −60 kN·m, MF = −30, MG = 0
Answer
Mark E: ME = −60 kN·m
7.Centroid & second moment of area
Answer: Ix = 98.8 × 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 = (30000·100 − 706.9·60)/(30000 − 706.9) = 101 mm
Full plate about ȳ
150·200³/12 + 30000·−0.9652² = 100 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·−40.97² = 1.226 × 10⁶ mm⁴
Subtract
Ix = 98.8 × 10⁶ mm⁴
Answer
Ix = 98.8 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σbottom = 99.19 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.
Split into two rectangles
Measure every y from the bottom edge.
flange: Af = 100·20 = 2000 mm², yf = 300 + 20/2 = 310 mm
web: Aw = 18·300 = 5400 mm², yw = 300/2 = 150 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (2000·310 + 5400·150)/7400 = 193.2 mm
Parallel-axis theorem
flange: 100·20³/12 + 2000·(310 − 193.2)² = 2.733 × 10⁷ mm⁴
web: 18·300³/12 + 5400·(150 − 193.2)² = 5.06 × 10⁷ mm⁴
Ix = 2.733 × 10⁷ + 5.06 × 10⁷ = 7.793 × 10⁷ mm⁴
Bending stress
The bottom edge is ȳ from the axis.
y = 193.2 mm
σ = M·y/I = 4 × 10⁷·193.2/7.793 × 10⁷ = 99.19 MPa
Answer
σbottom = 99.19 MPa
9.Mohr’s circle & stress transformation
Answer: τmax = 71.59 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 = √(((110 − −20)/2)² + 30²) = 71.59 MPa
Answer
τmax = 71.59 MPa
10.Mohr’s circle & stress transformation
Answer: σθ = 67.08 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 = 85 + −15·0.766 + 10·−0.6428 = 67.08 MPa
Answer
σθ = 67.08 MPa
11.Beam deflection
Answer: δ = 54.82 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 = 8 N/mm, L = 5000 mm
EI = 1.14 × 10¹³ N·mm²
Deflection
δ = wL⁴/(8EI) = 8·5000⁴/(8·1.14 × 10¹³) = 54.82 mm
Answer
δ = 54.82 mm
12.Beam deflection
Answer: RB = 20.31 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 = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)
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) = 5PL³/(48EI) → RB = 5P/16 = 5·65/16 = 20.31 kN
Answer
RB = 20.31 kN
13.Column buckling
Answer: Pcr = 77.24 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·306796/2800² = 77244 N = 77.24 kN
Answer
Pcr = 77.24 kN
14.Column buckling
Answer: d = 65.94 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 = 3·95000 = 285000 N
KL = 0.5·3000 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·285000·1500²/(π³·70000))^(1/4) = 65.94 mm
Answer
d = 65.94 mm
Check: the slenderness KL/r is 90.99, so the column is slender and Euler applies.
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