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 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaδ = 0.3 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 = 90 kNsegment 1: Ø25 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)
σmax = MPa
3.Torsion of shaftsMedium
A solid brass shaft 2.5 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.
T = 1500 N·mL = 2.5 mG = 39 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
A hollow shaft must transmit 50 kW at 750 rpm. 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.
P = 50 kWN = 750 rpmdi/do = 0.5τallow = 50 MPa
do = mm
5.Shear force & bending momentMedium
The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment Mx under the load P1. Use sagging moments positive and upward shear on the left face positive.
L = 7 mP1 = 60 kN at 4.5 mP2 = 45 kN at 7 msupports at x = 0 and x = 6 m
Mx = kN·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 = 6 mw = 8 kN/m from 3 to 5 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 = 160 mmtf = 10 mmtw = 10 mmhw = 100 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
A 200 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 40 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: 80 × 40 mmhole centre: y = 105 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
A point has σₓ = 110 MPa, σᵧ = −40 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 110 MPaσy = −40 MPaτxy = −50 MPa
θs = °
10.Mohr’s circle & stress transformationHard
A solid shaft of diameter 20 mm carries an axial compressive force of 140 kN together with a torque of 100 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 20 mmF = −140 kNT = 100 N·m
σ1 = 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 = 8 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction RB at the prop.
L = 5 mw = 8 kN/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 = 6 mE = 200 GPa (steel)b × h = 20 × 80 mm
Pcr = kN
14.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 6 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: P = 19.96 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 = 314.2 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(100000·706.9) = 7.074 × 10⁻⁶ mm/N
Load
P = 0.3/1.503 × 10⁻⁵ = 19958 N = 19.96 kN
Answer
P = 19.96 kN
2.Axial stress & strain
Answer: σmax = 795.8 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·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 183.3 MPa
σ2 = 795.8 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 795.8 MPa
Answer
σmax = 795.8 MPa
3.Torsion of shafts
Answer: d = 78.21 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.
Required polar moment
T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 2.5 m = 2500 mm
G = 39 GPa = 39000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 1.5 × 10⁶·2500/(39000·0.02618) = 3.673 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 78.21 mm
Answer
d = 78.21 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: do = 41.05 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·750/60 = 78.54 rad/s
T = P/ω = 50000/78.54 = 636.6 N·m
Outer diameter from the allowable stress
T = 636.6 N·m = 636620 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·636620/(π·50·0.9375)) = 41.05 mm
di = 0.5·41.05 = 20.52 mm
Answer
do = 41.05 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 = 33.75 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.
Support reactions
Equilibrium of the whole beam. The overhang load also acts about A.
ΣMA = 0: RB·6 = 60·4.5 + 45·7
RB = 585/6 = 97.5 kN
ΣFy = 0: RA = 105 − 97.5 = 7.5 kN
Cut under P1 and keep the left part
Mx = RA·4.5 = 7.5·4.5 = 33.75 kN·m
Answer
Mx = 33.75 kN·m
6.Shear force & bending moment
Answer: Mmax = 17.78 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 = 8·2 = 16 kN, acting at x̄ = 4 m
ΣMA = 0: RB·6 = 8·2·4
RB = 64/6 = 10.67 kN
ΣFy = 0: RA = 16 − 10.67 = 5.333 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 − 3) = 0
x0 = 3 + 5.333/8 = 3.667 m
Maximum moment
Mmax = RA·x0 − w·(x0 − 3)²/2 = 5.333·3.667 − 8·0.6667²/2 = 17.78 kN·m
Answer
Mmax = 17.78 kN·m
7.Centroid & second moment of area
Answer: Ix = 2.708 × 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 = 160·10 = 1600 mm², yf = 100 + 10/2 = 105 mm
web: Aw = 10·100 = 1000 mm², yw = 100/2 = 50 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (1600·105 + 1000·50)/2600 = 83.85 mm
Parallel-axis theorem
flange: 160·10³/12 + 1600·(105 − 83.85)² = 729310 mm⁴
web: 10·100³/12 + 1000·(50 − 83.85)² = 1.979 × 10⁶ mm⁴
Ix = 729310 + 1.979 × 10⁶ = 2.708 × 10⁶ mm⁴ = 2.708 × 10⁶ mm⁴
Answer
Ix = 2.708 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: Ix = 52.6 × 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 − 3200·105)/(30000 − 3200) = 71.42 mm
Full plate about ȳ
200·150³/12 + 30000·3.582² = 56.63 × 10⁶ mm⁴
Hole about ȳ
80·40³/12 + 3200·33.58² = 4.035 × 10⁶ mm⁴
Subtract
Ix = 52.6 × 10⁶ mm⁴
Answer
Ix = 52.6 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation
Answer: θs = −61.85 °
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.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −50/90.14 = −0.5547, sin 2θs = −75/90.14 = −0.8321
2θs = atan2(−75, −50) = −123.7°
Angle on the element
θs = −61.85°
check: θs − θp = −61.85 − (−16.85) = −45°
Answer
θs = −61.85°
Positive angles are counter-clockwise.
10.Mohr’s circle & stress transformation
Answer: σ1 = 8.916 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.
Stresses at the surface
σ = F/A = −140000/(π·20²/4) = −140000/314.2 = −445.6 MPa
τ = 16T/(πd³) = 16·100000/(π·20³) = 63.66 MPa
Mohr’s circle
C = σ/2 = −222.8 MPa
R = √((−222.8)² + 63.66²) = 231.7 MPa
σ1 = C + R = 8.916 MPa
σ2 = C − R = −454.6 MPa
Answer
σ1 = 8.916 MPa
11.Beam deflection
Answer: δmax = 63.49 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 = 8000 mm
EI = 8.4 × 10¹² N·mm²
Deflection
δmax = 5wL⁴/(384EI) = 5·10·8000⁴/(384·8.4 × 10¹²) = 63.49 mm
Answer
δmax = 63.49 mm
12.Beam deflection
Answer: RB = 15 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 = wL⁴/(8EI)
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) = wL⁴/(8EI) → RB = 3wL/8 = 3·8·5/8 = 15 kN
Answer
RB = 15 kN
13.Column buckling
Answer: Pcr = 5.968 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 = 80·20³/12 = 53333 mm⁴
Effective length
KL = 0.7·6000 = 4200 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·53333/4200² = 5968 N = 5.968 kN
Answer
Pcr = 5.968 kN
14.Column buckling
Answer: σcr = 7.676 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·6000 = 3000 mm
KL/r = 3000/10 = 300
Critical stress
σcr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa
Answer
σcr = 7.676 MPa
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