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 25 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 30 mm, length 400 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 400 mm, E = 110 GPaδ = 0.2 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 = 35 kNsegment 1: Ø35 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 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 = 250 N·md = 50 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 50 mm is fixed at A. A torque of 100 N·m acts at B, 2 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 50 mmTB = 100 N·mTC = 200 N·mAB = 2 mBC = 0.5 mG = 39 GPa
φC = °
5.Shear force & bending momentMedium
The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the shear force Vx just to the left of support B. Use sagging moments positive and upward shear on the left face positive.
L = 5.5 mP1 = 30 kN at 1 mP2 = 25 kN at 5.5 msupports at x = 0 and x = 4 m
Vx = kN
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 @ 0 m40 kN @ 12 m20 kN @ 8 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 symmetric I-section has flanges 100 mm wide and 24 mm thick, an overall depth of 240 mm and a web 8 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.
B = 100 mmH = 240 mmtf = 24 mmtw = 8 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 15 kN·m. Find the bending stress σbottom at the bottom edge.
bf = 220 mmtf = 15 mmtw = 10 mmhw = 160 mmM = 15 kN·m
σbottom = MPa
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 160 MPa, σᵧ = −20 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 160 MPaσy = −20 MPaτxy = 60 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = 0 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.
σx = 150 MPaσy = 0 MPaτxy = 20 MPaθ = 40°
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 = 12 kN/mL = 4 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δmax.
P = 4 kNL = 4 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δmax = 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 = 4 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
14.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: P = 21.69 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 = 490.9 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·490.9) = 4.074 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(110000·706.9) = 5.144 × 10⁻⁶ mm/N
Load
P = 0.2/9.219 × 10⁻⁶ = 21695 N = 21.69 kN
Answer
P = 21.69 kN
2.Axial stress & strain
Answer: σmax = 111.4 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·20²/4 = 314.2 mm²
σ1 = 36.38 MPa
σ2 = 111.4 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 111.4 MPa
Answer
σmax = 111.4 MPa
3.Torsion of shafts
Answer: φ = 0.7295 °
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) = 250·1000·2500/(80000·613592) = 0.01273 rad
φ = 0.01273·180/π = 0.7295°
Answer
φ = 0.7295°
4.Torsion of shafts
Answer: φC = 1.676 °
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 = 200 N·m
TAB = TC + TB = 200 + 100 = 300 N·m
Twist of each segment
J = π·50⁴/32 = 613592 mm⁴
G = 39 GPa = 39000 MPa
φAB = 300000·2000/(39000·613592) = 0.02507 rad
φBC = 200000·500/(39000·613592) = 0.004179 rad
Twist of end C
φC = 0.02507 + 0.004179 = 0.02925 rad → |φC| = 1.676°
Answer
φC = 1.676°
5.Shear force & bending moment
Answer: Vx = −16.88 kN
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. The overhang load also acts about A.
ΣMA = 0: RB·4 = 30·1 + 25·5.5
RB = 167.5/4 = 41.88 kN
ΣFy = 0: RA = 55 − 41.88 = 13.13 kN
Cut just left of B and keep the left part
Vx = RA − P1 = 13.13 − 30 = −16.88 kN
Answer
Vx = −16.88 kN
6.Shear force & bending moment
Answer: F (x = 10 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 = 20 kN, R2 = 60 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 = −40, MC = −40, MD = −40, ME = −40, MF = −80 kN·m, MG = 0
Answer
Mark F: MF = −80 kN·m
7.Centroid & second moment of area
Answer: Ix = 60.94 × 10⁶ mm⁴
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.
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 = 240 − 2·24 = 192 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (100·240³ − 92·192³)/12
Ix = 6.094 × 10⁷ mm⁴ = 60.94 × 10⁶ mm⁴
Answer
Ix = 60.94 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σbottom = 177.7 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 = 220·15 = 3300 mm², yf = 160 + 15/2 = 167.5 mm
web: Aw = 10·160 = 1600 mm², yw = 160/2 = 80 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (3300·167.5 + 1600·80)/4900 = 138.9 mm
Parallel-axis theorem
flange: 220·15³/12 + 3300·(167.5 − 138.9)² = 2.756 × 10⁶ mm⁴
web: 10·160³/12 + 1600·(80 − 138.9)² = 8.969 × 10⁶ mm⁴
Ix = 2.756 × 10⁶ + 8.969 × 10⁶ = 1.173 × 10⁷ mm⁴
Bending stress
The bottom edge is ȳ from the axis.
y = 138.9 mm
σ = M·y/I = 1.5 × 10⁷·138.9/1.173 × 10⁷ = 177.7 MPa
Answer
σbottom = 177.7 MPa
9.Mohr’s circle & stress transformation
Answer: τmax = 108.2 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 = √(((160 − −20)/2)² + 60²) = 108.2 MPa
Answer
τmax = 108.2 MPa
10.Mohr’s circle & stress transformation
Answer: σθ = 107.7 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 = 75 + 75·0.1736 + 20·0.9848 = 107.7 MPa
Answer
σθ = 107.7 MPa
11.Beam deflection
Answer: δmax = 23.53 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 = 12 N/mm, L = 4000 mm
EI = 1.7 × 10¹² N·mm²
Deflection
δmax = 5wL⁴/(384EI) = 5·12·4000⁴/(384·1.7 × 10¹²) = 23.53 mm
Answer
δmax = 23.53 mm
12.Beam deflection
Answer: δmax = 5.345 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 = 4000 N, L = 4000 mm
EI = 1.7 × 10¹² N·mm²
One load at L/3
δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 1333 mm
δ₁ = 2.672 mm
Add both loads
δmax = 2·δ₁ = 23PL³/(648EI) = 5.345 mm
Answer
δmax = 5.345 mm
13.Column buckling
Answer: Pcr = 4.905 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/4000² = 4905 N = 4.905 kN
Answer
Pcr = 4.905 kN
14.Column buckling
Answer: σcr = 197.4 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 1·2000 = 2000 mm
KL/r = 2000/20 = 100
Critical stress
σcr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa
Answer
σcr = 197.4 MPa
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