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 axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 25 kNd = 25 mmL = 2.5 mE = 100 GPa
ε = µε
2.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 300 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 300 mm, E = 120 GPaP = 50 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 = 200 N·md = 20 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 1500 N·m is applied at C, 2.5 m from A and 0.5 m from B. Find the angle of rotation φC of section C, in degrees.
d = 50 mmT0 = 1500 N·ma = 2.5 mb = 0.5 mG = 39 GPa
φC = °
5.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 20 kN·m at x = 3 m from A, and no other load. Find the bending moment Mx just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.
L = 5 mM0 = 20 kN·m clockwise, at 3 m
Mx = 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.
20 kN @ 0 m30 kN @ 6 m8 kN/m, 1–5 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
A 150 mm × 240 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 168 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 240 mmd = 60 mmhole centre: y = 168 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 20 kN·m. Find the size of the bending stress σtop at the top edge.
bf = 260 mmtf = 30 mmtw = 8 mmhw = 240 mmM = 20 kN·m
σtop = MPa
9.Mohr’s circle & stress transformationMedium
A point has σₓ = 20 MPa, σᵧ = −10 MPa and τₓᵧ = −20 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 = 20 MPaσy = −10 MPaτxy = −20 MPa
θs = °
10.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries a bending moment of 750 N·m together with a torque of 2000 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 50 mmM = 750 N·mT = 2000 N·m
σ1 = 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 = 3 mE = 70 GPa (aluminium)I = 120 × 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 size of the bending moment MA at the fixed end.
L = 8 mP = 75 kN at 4 m
MA = kN·m
13.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 70 GPa (aluminium)d = 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 = 30 mm
σcr = MPa
3Solutions
1.Axial stress & strain
Answer: ε = 509.3 µε
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·25²/4 = 490.9 mm²
σ = P/A = 25·1000/490.9 = 50.93 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 50.93/100000 = 0.0005093 = 509.3 µε
Answer
ε = 509.3 µε
2.Axial stress & strain
Answer: σ2 = 41.75 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 120000·706.9/300 = 282743 N/mm
Share of segment 2
F2 = 50000·282743/479093 = 29508 N
Stress
σ2 = 29508/706.9 = 41.75 MPa (compression)
Answer
σ2 = 41.75 MPa
3.Torsion of shafts
Answer: φ = 22.45 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·800/(26000·15708) = 0.3918 rad
φ = 0.3918·180/π = 22.45°
Answer
φ = 22.45°
4.Torsion of shafts
Answer: φC = 1.496 °
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 1500
TA·a = TB·b
TA = 1500·0.5/3 = 250 N·m
TB = 1500·2.5/3 = 1250 N·m
Rotation of C
Use segment AC: it is fixed at A and carries TA.
J = π·50⁴/32 = 613592 mm⁴
G = 39 GPa = 39000 MPa
φC = TA·a/(GJ) = 250000·2500/(39000·613592) = 0.02612 rad = 1.496°
Answer
φC = 1.496°
5.Shear force & bending moment
Answer: Mx = −12 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. An applied couple makes the bending moment diagram jump by the size of the couple.
Support reactions
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
RB = M0/L = 20/5 = 4 kN (upward)
RA = −RB = −4 kN (downward)
Cut just left of the couple
The left part contains only RA.
Mx = RA·a = −4·3 = −12 kN·m
Answer
Mx = −12 kN·m
6.Shear force & bending moment
Answer: F (x = 5 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 = 33.5 kN, R2 = 48.5 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 = −20, MC = −10.5, MD = −9, ME = −15.5, MF = −30 kN·m, MG = 0
Answer
Mark F: MF = −30 kN·m
7.Centroid & second moment of area
Answer: Ix = 165.1 × 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 = (36000·120 − 2827·168)/(36000 − 2827) = 115.9 mm
Full plate about ȳ
150·240³/12 + 36000·4.091² = 173.4 × 10⁶ mm⁴
Hole about ȳ
π·60⁴/64 + 2827·52.09² = 8.308 × 10⁶ mm⁴
Subtract
Ix = 165.1 × 10⁶ mm⁴
Answer
Ix = 165.1 × 10⁶ mm⁴
8.Centroid & second moment of area
Answer: σtop = 22 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 = 260·30 = 7800 mm², yf = 240 + 30/2 = 255 mm
web: Aw = 8·240 = 1920 mm², yw = 240/2 = 120 mm
Centroid
ȳ = (Afyf + Awyw)/(Af + Aw) = (7800·255 + 1920·120)/9720 = 228.3 mm
Parallel-axis theorem
flange: 260·30³/12 + 7800·(255 − 228.3)² = 6.132 × 10⁶ mm⁴
web: 8·240³/12 + 1920·(120 − 228.3)² = 3.175 × 10⁷ mm⁴
Ix = 6.132 × 10⁶ + 3.175 × 10⁷ = 3.788 × 10⁷ mm⁴
Bending stress
The top edge is h − ȳ from the axis.
y = 270 − 228.3 = 41.67 mm
σ = M·y/I = 2 × 10⁷·41.67/3.788 × 10⁷ = 22 MPa
Answer
σtop = 22 MPa
9.Mohr’s circle & stress transformation
Answer: θs = −71.57 °
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 = −20/25 = −0.8, sin 2θs = −15/25 = −0.6
2θs = atan2(−15, −20) = −143.1°
Angle on the element
θs = −71.57°
check: θs − θp = −71.57 − (−26.57) = −45°
Answer
θs = −71.57°
Positive angles are counter-clockwise.
10.Mohr’s circle & stress transformation
Answer: σ1 = 117.6 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 = 32M/(πd³) = 32·750000/(π·50³) = 61.12 MPa
τ = 16T/(πd³) = 16·2000000/(π·50³) = 81.49 MPa
Mohr’s circle
C = σ/2 = 30.56 MPa
R = √(30.56² + 81.49²) = 87.03 MPa
σ1 = C + R = 117.6 MPa
σ2 = C − R = −56.47 MPa
Answer
σ1 = 117.6 MPa
11.Beam deflection
Answer: δ = 14.46 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 = 3000 mm
EI = 8.4 × 10¹² N·mm²
Deflection
δ = wL⁴/(8EI) = 12·3000⁴/(8·8.4 × 10¹²) = 14.46 mm
Answer
δ = 14.46 mm
12.Beam deflection
Answer: MA = 112.5 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 = 300 − 187.5 = 112.5 kN·m
Answer
MA = 112.5 kN·m
13.Column buckling
Answer: Pcr = 9.646 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 2·6000 = 12000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/12000² = 9646 N = 9.646 kN
Answer
Pcr = 9.646 kN
14.Column buckling
Answer: σcr = 4.318 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·6000 = 3000 mm
KL/r = 3000/7.5 = 400
Critical stress
σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa
Answer
σcr = 4.318 MPa
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