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 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.
P = 200 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 400 mm
δ = mm
2.Axial stress & strainHard
A steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 300 mm, E = 120 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, Ø30 mm, 600 mm, E = 200 GPasegment 2: copper, Ø40 mm, 300 mm, E = 120 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 = 30 mmL = 2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 50 kW at 1800 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.
P = 50 kWN = 1800 rpmdi/do = 0.6τallow = 40 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 2 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 MB at support B. Use sagging moments positive and upward shear on the left face positive.
L = 8 mP1 = 70 kN at 1.5 mP2 = 50 kN at 8 msupports at x = 0 and x = 6 m
MB = 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.
15 kN @ 0 m10 kN @ 12 m10 kN @ 4 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 hollow rectangular (box) section has outer width B = 240 mm, outer height H = 275 mm and a uniform wall thickness t = 20 mm. Find its radius of gyration r about the horizontal centroidal axis.
B = 240 mmH = 275 mmt = 20 mm
r = mm
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 bending stress σbottom at the bottom edge.
bf = 120 mmtf = 30 mmtw = 20 mmhw = 140 mmM = 10 kN·m
σbottom = MPa
9.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −40 MPa, σᵧ = 60 MPa and τₓᵧ = 50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −40 MPaσy = 60 MPaτxy = 50 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 0 MPa, σᵧ = 0 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 0 MPaσy = 0 MPaτxy = 30 MPaθ = -20°
τθ = MPa
11.Beam deflectionMedium
A simply supported beam carries a point load P at a = 2.5 m from support A. Find the deflection δP under the load.
P = 21 kNa = 2.5 mL = 6 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴
δP = mm
12.Beam deflectionHard
A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.
w = 6 kN/mL = 6 mE = 70 GPa (aluminium)
Ireq = × 10⁶ mm⁴
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 = 5 mE = 200 GPa (steel)b × h = 20 × 80 mm
Pcr = kN
14.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 6 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: δ = 0.1072 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·25²/4 = 490.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N
Shortening
δ = 200000·400/7.461 × 10⁸ = 0.1072 mm
Answer
δ = 0.1072 mm
2.Axial stress & strain
Answer: σ2 = 16.25 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 = 200000·706.9/600 = 235619 N/mm
k₂ = 120000·1257/300 = 502655 N/mm
Share of segment 2
F2 = 30000·502655/738274 = 20426 N
Stress
σ2 = 20426/1257 = 16.25 MPa (compression)
Answer
σ2 = 16.25 MPa
3.Torsion of shafts
Answer: φ = 22.17 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·2000/(26000·79522) = 0.3869 rad
φ = 0.3869·180/π = 22.17°
Answer
φ = 22.17°
4.Torsion of shafts
Answer: do = 33.85 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·1800/60 = 188.5 rad/s
T = P/ω = 50000/188.5 = 265.3 N·m
Outer diameter from the allowable stress
T = 265.3 N·m = 265258 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·265258/(π·40·0.8704)) = 33.85 mm
di = 0.6·33.85 = 20.31 mm
Answer
do = 33.85 mm
A solid shaft would need d = 32.32 mm; the hollow one is a little larger but much lighter.
5.Shear force & bending moment
Answer: MB = −100 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 = 70·1.5 + 50·8
RB = 505/6 = 84.17 kN
ΣFy = 0: RA = 120 − 84.17 = 35.83 kN
Cut at B and keep the overhang
Only P2 acts on the overhang, with lever arm a. It bends the beam upward over the support (hogging).
MB = −P2·a = −50·2 = −100 kN·m
Answer
MB = −100 kN·m
6.Shear force & bending moment
Answer: B (x = 2 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 = 23.75 kN, R2 = 11.25 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 = −30 kN·m, MC = −12.5, MD = −15, ME = −17.5, MF = −20, MG = 0
Answer
Mark B: MB = −30 kN·m
7.Centroid & second moment of area
Answer: r = 102.5 mm
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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.
Net area and second moment of area
b = 200 mm, h = 235 mm
A = BH − bh = 240·275 − 200·235 = 19000 mm²
I = (BH³ − bh³)/12 = 1.996 × 10⁸ mm⁴
Radius of gyration
r = √(I/A) = √(1.996 × 10⁸/19000) = 102.5 mm
Answer
r = 102.5 mm
8.Centroid & second moment of area
Answer: σbottom = 72.62 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 = 120·30 = 3600 mm², yf = 140 + 30/2 = 155 mm
web: Aw = 20·140 = 2800 mm², yw = 140/2 = 70 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (3600·155 + 2800·70)/6400 = 117.8 mm
Parallel-axis theorem
flange: 120·30³/12 + 3600·(155 − 117.8)² = 5.248 × 10⁶ mm⁴
web: 20·140³/12 + 2800·(70 − 117.8)² = 1.097 × 10⁷ mm⁴
Ix = 5.248 × 10⁶ + 1.097 × 10⁷ = 1.622 × 10⁷ mm⁴
Bending stress
The bottom edge is ȳ from the axis.
y = 117.8 mm
σ = M·y/I = 1 × 10⁷·117.8/1.622 × 10⁷ = 72.62 MPa
Answer
σbottom = 72.62 MPa
9.Mohr’s circle & stress transformation
Answer: τmax = 70.71 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 = √(((−40 − 60)/2)² + 50²) = 70.71 MPa
Answer
τmax = 70.71 MPa
10.Mohr’s circle & stress transformation
Answer: τθ = 22.98 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
τθ = −0·−0.6428 + 30·0.766 = 22.98 MPa
Answer
τθ = 22.98 MPa
11.Beam deflection
Answer: δP = 5.317 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 = 21000 N, a = 2500 mm, b = 3500 mm, L = 6000 mm
EI = 1.68 × 10¹³ N·mm²
Deflection under the load
δP = Pa²b²/(3EIL) = 21000·2500²·3500²/(3·1.68 × 10¹³·6000) = 5.317 mm
Answer
δP = 5.317 mm
12.Beam deflection
Answer: Ireq = 86.79 × 10⁶ mm⁴
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).
Deflection limit
δallow = L/360 = 6000/360 = 16.67 mm
Required second moment of area
I = 5wL⁴/(384·E·δallow) = 5·6·6000⁴/(384·70000·16.67) = 8.679 × 10⁷ mm⁴
Ireq = 86.79 × 10⁶ mm⁴
Answer
Ireq = 86.79 × 10⁶ mm⁴
13.Column buckling
Answer: Pcr = 8.594 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 = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·53333/3500² = 8594 N = 8.594 kN
Answer
Pcr = 8.594 kN
14.Column buckling
Answer: σcr = 17.48 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.7·6000 = 4200 mm
KL/r = 4200/12.5 = 336
Critical stress
σcr = π²E/(KL/r)² = π²·200000/336² = 17.48 MPa
Answer
σcr = 17.48 MPa
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