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 aluminium bar under the axial load.
P = 95 kNd = 12 mmL = 0.5 mE = 70 GPa
δ = mm
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 = 25 kNsegment 1: Ø35 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 400 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
T = 400 N·mdi/do = 0.75τallow = 75 MPa
do = mm
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 10 kWN = 500 rpmτallow = 80 MPa
d = mm
5.Shear force & bending momentMedium
The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.
L = 5 mP1 = 80 kN at 3 mP2 = 50 kN at 3.5 m
Mmax = 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 @ 6 m35 kN @ 3 m12 kN/m, 0–4 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
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 120 mmtf = 20 mmtw = 20 mmhw = 220 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 10 kN·m. Find the size of the bending stress σtop at the top edge.
bf = 280 mmtf = 10 mmtw = 10 mmhw = 260 mmM = 10 kN·m
σtop = MPa
9.Mohr’s circle & stress transformationMedium
A point has σₓ = 20 MPa, σᵧ = −50 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 20 MPaσy = −50 MPaτxy = −10 MPaθ = 35°
σy′ = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −20 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = −20 MPaσy = −60 MPaτxy = 30 MPaθ = 30°
σθ = 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 = 4 kN/mL = 6 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴
δmax = 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 size of the bending moment MA at the fixed end.
L = 5 mP = 75 kN at 2.5 m
MA = kN·m
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 = 5 mE = 200 GPa (steel)b × h = 40 × 100 mm
Pcr = kN
14.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 15 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
3Solutions
1.Axial stress & strain
Answer: δ = 6 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·12²/4 = 113.1 mm²
σ = P/A = 95·1000/113.1 = 840 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 840/70000 = 0.012 = 12000 µε
Elongation
δ = ε·L = 0.012·500 mm = 6 mm
Answer
δ = 6 mm
2.Axial stress & strain
Answer: σmax = 124.3 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·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 25.98 MPa
σ2 = 124.3 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 124.3 MPa
Answer
σmax = 124.3 MPa
3.Torsion of shafts
Answer: do = 34.12 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 = 400 N·m = 400000 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·400000/(π·75·0.6836)) = 34.12 mm
di = 0.75·34.12 = 25.59 mm
Answer
do = 34.12 mm
A solid shaft would need d = 30.06 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: d = 22.99 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·500/60 = 52.36 rad/s
T = P/ω = 10000/52.36 = 191 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·80)) = 22.99 mm
Answer
d = 22.99 mm
In practice you would round up to the next standard size.
5.Shear force & bending moment
Answer: Mmax = 141 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.
Support reactions
Equilibrium of the whole beam.
ΣMA = 0: RB·5 = 80·3 + 50·3.5
RB = 415/5 = 83 kN
ΣFy = 0: RA = 130 − 83 = 47 kN
Moment under each load
Keep the part left of each load.
under P1: M = RA·3 = 47·3 = 141 kN·m
under P2: M = RB·1.5 = 83·1.5 = 124.5 kN·m
Largest value
Mmax = 141 kN·m (under P1)
Answer
Mmax = 141 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 = 2.75 kN, R2 = 140.3 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 = −3.25, MC = −18.5, MD = −45.75, ME = −120 kN·m, MF = −60, MG = 0
Answer
Mark E: ME = −120 kN·m
7.Centroid & second moment of area
Answer: Ix = 40.19 × 10⁶ mm⁴
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 = 120·20 = 2400 mm², yf = 220 + 20/2 = 230 mm
web: Aw = 20·220 = 4400 mm², yw = 220/2 = 110 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (2400·230 + 4400·110)/6800 = 152.4 mm
Parallel-axis theorem
flange: 120·20³/12 + 2400·(230 − 152.4)² = 1.455 × 10⁷ mm⁴
web: 20·220³/12 + 4400·(110 − 152.4)² = 2.564 × 10⁷ mm⁴
Ix = 1.455 × 10⁷ + 2.564 × 10⁷ = 4.019 × 10⁷ mm⁴ = 40.19 × 10⁶ mm⁴
Answer
Ix = 40.19 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σtop = 17.84 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 = 280·10 = 2800 mm², yf = 260 + 10/2 = 265 mm
web: Aw = 10·260 = 2600 mm², yw = 260/2 = 130 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (2800·265 + 2600·130)/5400 = 200 mm
Parallel-axis theorem
flange: 280·10³/12 + 2800·(265 − 200)² = 1.185 × 10⁷ mm⁴
web: 10·260³/12 + 2600·(130 − 200)² = 2.739 × 10⁷ mm⁴
Ix = 1.185 × 10⁷ + 2.739 × 10⁷ = 3.924 × 10⁷ mm⁴
Bending stress
The top edge is h − ȳ from the axis.
y = 270 − 200 = 70 mm
σ = M·y/I = 1 × 10⁷·70/3.924 × 10⁷ = 17.84 MPa
Answer
σtop = 17.84 MPa
9.Mohr’s circle & stress transformation
Answer: σy′ = −17.57 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 for the y′ face
σy′ = −15 − 35·0.342 − (−10)·0.9397 = −17.57 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −12.43 MPa, σx′ + σy′ = −30 = σx + σy
Answer
σy′ = −17.57 MPa
10.Mohr’s circle & stress transformation
Answer: σθ = −4.019 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 = −40 + 20·0.5 + 30·0.866 = −4.019 MPa
Answer
σθ = −4.019 MPa
11.Beam deflection
Answer: δmax = 38.35 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 = 4 N/mm, L = 6000 mm
EI = 1.76 × 10¹² N·mm²
Deflection
δmax = 5wL⁴/(384EI) = 5·4·6000⁴/(384·1.76 × 10¹²) = 38.35 mm
Answer
δmax = 38.35 mm
12.Beam deflection
Answer: MA = 70.31 kN·m
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 = 5·75/16 = 23.44 kN
Moment at the fixed end
Moments about A of the loads and the prop force.
MA = P·L/2 − RB·L = 187.5 − 117.2 = 70.31 kN·m
Answer
MA = 70.31 kN·m
13.Column buckling
Answer: Pcr = 168.4 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 = 100·40³/12 = 533333 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·533333/2500² = 168441 N = 168.4 kN
Answer
Pcr = 168.4 kN
14.Column buckling
Answer: d = 54.26 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 = 2.5·15000 = 37500 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·2800²/(π³·70000))^(1/4) = 54.26 mm
Answer
d = 54.26 mm
Check: the slenderness KL/r is 206.4, so the column is slender and Euler applies.
ExamProblems · examproblems.com/courses/mechanics-of-materials