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 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 200 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 200 mm, E = 110 GPaδ = 0.2 mm
P = kN
2.Axial stress & strainHard
A steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaP = 30 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 = 400 N·md = 25 mmL = 1 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 300 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 10 kWN = 300 rpmτallow = 50 MPa
d = mm
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.
45 kN @ 6 m6 kN/m, 2–8 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)
6.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 8 mw = 12 kN/m from 0 to 6 m
Mmax = kN·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 = 100 mmtf = 25 mmtw = 12 mmhw = 100 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A symmetric I-section has flanges 160 mm wide and 18 mm thick, an overall depth of 240 mm and a web 10 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the maximum bending stress σmax in the section.
B = 160 mmH = 240 mmtf = 18 mmtw = 10 mmM = 80 kN·m
σmax = MPa
9.Mohr’s circle & stress transformationMedium
A point has σₓ = −20 MPa and σᵧ = 60 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 115 MPa?
σx = −20 MPaσy = 60 MPaσ1 = 115 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 12 mm holds gas at a gauge pressure of 2 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 500 mmt = 12 mmp = 2 MPaα = 55°
σn = 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 = 12 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 160 × 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 = 50 kN at 1.5 m
RB = kN
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 = 3 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
14.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 25 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 25 kNn = 2.5L = 4 mE = 200 GPa (steel)
d = mm
3Solutions
1.Axial stress & strain
Answer: P = 65.82 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 = 1257 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(110000·1257) = 1.447 × 10⁻⁶ mm/N
Load
P = 0.2/3.038 × 10⁻⁶ = 65824 N = 65.82 kN
Answer
P = 65.82 kN
2.Axial stress & strain
Answer: σ2 = 10.93 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 = 200000·490.9/300 = 327249 N/mm
k₂ = 110000·1257/500 = 276460 N/mm
Share of segment 2
F2 = 30000·276460/603709 = 13738 N
Stress
σ2 = 13738/1257 = 10.93 MPa (compression)
Answer
σ2 = 10.93 MPa
3.Torsion of shafts
Answer: φ = 22.99 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·1000/(26000·38350) = 0.4012 rad
φ = 0.4012·180/π = 22.99°
Answer
φ = 22.99°
4.Torsion of shafts
Answer: d = 31.89 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 = 10000/31.42 = 318.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·318310/(π·50)) = 31.89 mm
Answer
d = 31.89 mm
In practice you would round up to the next standard size.
5.Shear force & bending moment
Answer: D (x = 6 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 = 43.5 kN, R2 = 37.5 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 = 87, MC = 162, MD = 213 kN·m, ME = 150, MF = 75, MG = 0
Answer
Mark D: MD = 213 kN·m
6.Shear force & bending moment
Answer: Mmax = 84.38 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.
UDL resultant: W = w·ℓ = 12·6 = 72 kN, acting at x̄ = 3 m
ΣMA = 0: RB·8 = 12·6·3
RB = 216/8 = 27 kN
ΣFy = 0: RA = 72 − 27 = 45 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
RA − w·(x0 − 0) = 0
x0 = 0 + 45/12 = 3.75 m
Maximum moment
Mmax = RA·x0 − w·(x0 − 0)²/2 = 45·3.75 − 12·3.75²/2 = 84.38 kN·m
Answer
Mmax = 84.38 kN·m
7.Centroid & second moment of area
Answer: Ix = 4.297 × 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 = 100·25 = 2500 mm², yf = 100 + 25/2 = 112.5 mm
web: Aw = 12·100 = 1200 mm², yw = 100/2 = 50 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (2500·112.5 + 1200·50)/3700 = 92.23 mm
Parallel-axis theorem
flange: 100·25³/12 + 2500·(112.5 − 92.23)² = 1.157 × 10⁶ mm⁴
web: 12·100³/12 + 1200·(50 − 92.23)² = 3.14 × 10⁶ mm⁴
Ix = 1.157 × 10⁶ + 3.14 × 10⁶ = 4.297 × 10⁶ mm⁴ = 4.297 × 10⁶ mm⁴
Answer
Ix = 4.297 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σmax = 122.8 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.
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 − 2·18 = 204 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (160·240³ − 150·204³)/12
Ix = 7.82 × 10⁷ mm⁴
Bending stress
y = 120 mm
σ = M·y/I = 8 × 10⁷·120/7.82 × 10⁷ = 122.8 MPa
Answer
σmax = 122.8 MPa
9.Mohr’s circle & stress transformation
Answer: τxy = 86.17 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.
Required radius
C = (−20 + 60)/2 = 20 MPa
R = σ1 − C = 115 − 20 = 95 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(95² − 40²) = 86.17 MPa
Answer
τxy = 86.17 MPa
10.Mohr’s circle & stress transformation
Answer: σn = 27.69 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 500/2 = 250 mm
σh = pr/t = 2·250/12 = 41.67 MPa
σa = pr/(2t) = 20.83 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 20.83·sin²55° + 41.67·cos²55° = 27.69 MPa
Answer
σn = 27.69 MPa
11.Beam deflection
Answer: δ = 5.232 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 = 2500 mm
EI = 1.12 × 10¹³ N·mm²
Deflection
δ = wL⁴/(8EI) = 12·2500⁴/(8·1.12 × 10¹³) = 5.232 mm
Answer
δ = 5.232 mm
12.Beam deflection
Answer: RB = 15.63 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 = 5·50/16 = 15.63 kN
Answer
RB = 15.63 kN
13.Column buckling
Answer: Pcr = 374.3 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 = 80·40³/12 = 426667 mm⁴
Effective length
KL = 0.5·3000 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·426667/1500² = 374314 N = 374.3 kN
Answer
Pcr = 374.3 kN
14.Column buckling
Answer: d = 80.16 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 = 2.5·25000 = 62500 N
KL = 2·4000 = 8000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·62500·8000²/(π³·200000))^(1/4) = 80.16 mm
Answer
d = 80.16 mm
Check: the slenderness KL/r is 399.2, so the column is slender and Euler applies.
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