Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. 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

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. 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

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. 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
Ø 16 mmAluminium0.5 mP = 30 kN

Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).

P = 30 kNd = 16 mmL = 0.5 mE = 70 GPa
ε = µε
2.Axial stress & strainHard
100 kN12400 mm500 mmØ20Ø30σ2​ = ?

A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 500 mm, E = 110 GPaP = 100 kN
σ2 = MPa
3.Torsion of shaftsMedium
d1​ = 75 mmd2​ = 40 mmT = 1000 N·mABC1 m2.5 m

A stepped solid brass shaft is fixed at A. Segment AB has diameter 75 mm and length 1 m; segment BC has diameter 40 mm and length 2.5 m. A torque of 1000 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.

T = 1000 N·md1 = 75 mm, L1 = 1 md2 = 40 mm, L2 = 2.5 m
τmax = MPa
4.Torsion of shaftsHard
Tdo​ = 60 mm, di​ = 48 mm

A hollow shaft transmits 50 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.

P = 50 kWN = 500 rpmdo = 60 mmdi = 48 mm
τmax = MPa
5.Shear force & bending momentMedium
10 kN/m20 kNABCDEFG1 m1 m1 m1 m1 m1 m

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 @ 5 m10 kN/m, 0–4 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
6.Shear force & bending momentHard
w = 4 kN/mP = 35 kNABsection1 m2.5 m1.5 m

Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 35 kN at 3.5 mw = 4 kN/m from 0 to 5 mx = 1 m
Mx = kN·m
7.Centroid & second moment of areaMedium
xD = 230d = 206dimensions in mm

A circular tube has outside diameter D = 230 mm and inside diameter d = 206 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 230 mmd = 206 mm
r = mm
8.Centroid & second moment of areaHard
100 mm240 mm168 mm30 × 30xIx​ = ?

A 100 mm × 240 mm plate has a rectangular cut-out 30 mm wide and 30 mm high, centred on the 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 = 100 mmh = 240 mmcut-out: 30 × 30 mmhole centre: y = 168 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 100σy​ = 100MPaτxy​ = ?

A point has σₓ = 100 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 125 MPa?

σx = 100 MPaσy = 100 MPaσ1 = 125 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
40°D = 1200 mm, t = 6 mm, p = 0.8 MPap

A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 6 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 40° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.

D = 1200 mmt = 6 mmp = 0.8 MPaα = 40°
σn = MPa
11.Beam deflectionMedium
w = 20 kN/mδL = 2.5 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.

w = 20 kN/mL = 2.5 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
w = 15 kN/mAB5 m

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 size of the bending moment MA at the fixed end.

L = 5 mw = 15 kN/m
MA = kN·m
13.Column bucklingMedium
PL = 2.5 m20 × 100 mmcross-section

A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 2.5 mE = 200 GPa (steel)b × h = 20 × 100 mm
Pcr = kN
14.Column bucklingHard
P = 140 kNL = 2 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 140 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 140 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 2132 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·16²/4 = 201.1 mm²
σ = P/A = 30·1000/201.1 = 149.2 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 149.2/70000 = 0.002132 = 2132 µε

Answer

ε = 2132 µε
2.Axial stress & strain

Answer: σ2 = 70.38 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 = 200000·314.2/400 = 157080 N/mm
k₂ = 110000·706.9/500 = 155509 N/mm

Share of segment 2

F2 = 100000·155509/312588 = 49749 N

Stress

σ2 = 49749/706.9 = 70.38 MPa (compression)

Answer

σ2 = 70.38 MPa
3.Torsion of shafts

Answer: τmax = 79.58 MPa

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.

Stress in each segment

T = 1000 N·m = 1 × 10⁶ N·mm
τAB = 16·1 × 10⁶/(π·75³) = 12.07 MPa
τBC = 16·1 × 10⁶/(π·40³) = 79.58 MPa

Answer

τmax = 79.58 MPa
4.Torsion of shafts

Answer: τmax = 38.14 MPa

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·500/60 = 52.36 rad/s
T = P/ω = 50000/52.36 = 954.9 N·m

Polar second moment of area

J = π(60⁴ − 48⁴)/32 = 751193 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 954930·30/751193 = 38.14 MPa

Answer

τmax = 38.14 MPa
5.Shear force & bending moment

Answer: D (x = 3 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 = 30 kN, R2 = 30 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 = 25, MC = 40, MD = 45 kN·m, ME = 40, MF = 30, MG = 0

Answer

Mark D: MD = 45 kN·m
6.Shear force & bending moment

Answer: Mx = 18.5 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.

UDL resultant: W = w·ℓ = 4·5 = 20 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 35·3.5 + 4·5·2.5
RB = 172.5/5 = 34.5 kN
ΣFy = 0: RA = 55 − 34.5 = 20.5 kN
w = 4 kN/mP = 35 kNABRA​ = 20.5RB​ = 34.5section1 m2.5 m1.5 m

Cut at x = 1 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 20.5·1 − 4·1·0.5 = 18.5 kN·m

Answer

Mx = 18.5 kN·m
7.Centroid & second moment of area

Answer: r = 77.19 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

A = π(D² − d²)/4 = 8218 mm²
I = π(D⁴ − d⁴)/64 = 4.897 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(4.897 × 10⁷/8218) = 77.19 mm

Answer

r = 77.19 mm
8.Centroid & second moment of area

Answer: Ix = 113 × 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 = (24000·120 − 900·168)/(24000 − 900) = 118.1 mm

Full plate about ȳ

100·240³/12 + 24000·1.87² = 115.3 × 10⁶ mm⁴

Hole about ȳ

30·30³/12 + 900·49.87² = 2.306 × 10⁶ mm⁴

Subtract

Ix = 113 × 10⁶ mm⁴

Answer

Ix = 113 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τxy = 25 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 = (100 + 100)/2 = 100 MPa
R = σ1 − C = 125 − 100 = 25 MPa
σxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(25² − 0²) = 25 MPa

Answer

τxy = 25 MPa
10.Mohr’s circle & stress transformation

Answer: σn = 63.47 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 = 1200/2 = 600 mm
σh = pr/t = 0.8·600/6 = 80 MPa
σa = pr/(2t) = 40 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 = 40·sin²40° + 80·cos²40° = 63.47 MPa

Answer

σn = 63.47 MPa
11.Beam deflection

Answer: δmax = 2.211 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 = 20 N/mm, L = 2500 mm
EI = 4.6 × 10¹² N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·20·2500⁴/(384·4.6 × 10¹²) = 2.211 mm

Answer

δmax = 2.211 mm
12.Beam deflection

Answer: MA = 46.88 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 = 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 = 3·15·5/8 = 28.13 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

MA = wL²/2 − RB·L = 187.5 − 140.6 = 46.88 kN·m

Answer

MA = 46.88 kN·m
13.Column buckling

Answer: Pcr = 84.22 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 = 100·20³/12 = 66667 mm⁴

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·66667/1250² = 84221 N = 84.22 kN

Answer

Pcr = 84.22 kN
14.Column buckling

Answer: d = 61.65 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 = 2.5·140000 = 350000 N
KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·350000·2000²/(π³·200000))^(1/4) = 61.65 mm

Answer

d = 61.65 mm

Check: the slenderness KL/r is 129.8, so the column is slender and Euler applies.

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