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
300 kNØ40110 GPaØ80/60600 mmδ = ?

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 110 GPaL = 600 mm
δ = mm
2.Axial stress & strainHard
50 kN12400 mm200 mmØ30Ø40σ2​ = ?

A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 200 mm, E = 70 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, Ø30 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 200 mm, E = 70 GPaP = 50 kN
σ2 = MPa
3.Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 2.5 mφ = ?

A solid brass 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 = 750 N·md = 40 mmL = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 50 mm, di​ = 38 mm

A hollow shaft transmits 5 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.

P = 5 kWN = 750 rpmdo = 50 mmdi = 38 mm
τmax = MPa
5.Shear force & bending momentMedium
12 kN/m25 kNABCDEFG2 m2 m2 m2 m2 m2 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.

25 kN @ 6 m12 kN/m, 0–10 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
6.Shear force & bending momentHard
20 kN15 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 @ 6 m15 kN @ 2 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)
7.Centroid & second moment of areaMedium
150 mm240 mm72 mmØ60xIx​ = ?

A 150 mm × 240 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 72 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 = 72 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
200 mm240 mm84 mm60 × 40xIx​ = ?

A 200 mm × 240 mm plate has a rectangular cut-out 60 mm wide and 40 mm high, centred on the vertical centre line with its centre 84 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 200 mmh = 240 mmcut-out: 60 × 40 mmhole centre: y = 84 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 120σy​ = 100τxy​ = −40MPa

A point in a loaded part has σₓ = 120 MPa, σᵧ = 100 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 120 MPaσy = 100 MPaτxy = −40 MPa
τmax = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 90σy​ = 40τxy​ = 40MPa

A point in a loaded part has σₓ = 90 MPa, σᵧ = 40 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σx = 90 MPaσy = 40 MPaτxy = 40 MPaθ = 40°
σθ = MPa
11.Beam deflectionMedium
w = 4 kN/mABθ2.5 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.

w = 4 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 12 × 10⁶ mm⁴
θA = rad
12.Beam deflectionHard
w = 15 kN/mAB3 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 = 3 mw = 15 kN/m
MA = kN·m
13.Column bucklingMedium
PL = 2.5 m20 × 50 mmcross-section

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 = 2.5 mE = 200 GPa (steel)b × h = 20 × 50 mm
Pcr = kN
14.Column bucklingHard
P = 30 kNL = 4 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 30 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 30 kNn = 3L = 4 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.3649 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·40²/4 = 1257 mm²
At = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·2199 = 4.932 × 10⁸ N

Shortening

δ = 300000·600/4.932 × 10⁸ = 0.3649 mm

Answer

δ = 0.3649 mm
2.Axial stress & strain

Answer: σ2 = 22.06 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 = 200000·706.9/400 = 353429 N/mm
k₂ = 70000·1257/200 = 439823 N/mm

Share of segment 2

F2 = 50000·439823/793252 = 27723 N

Stress

σ2 = 27723/1257 = 22.06 MPa (compression)

Answer

σ2 = 22.06 MPa
3.Torsion of shafts

Answer: φ = 10.96 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·2500/(39000·251327) = 0.1913 rad
φ = 0.1913·180/π = 10.96°

Answer

φ = 10.96°
4.Torsion of shafts

Answer: τmax = 3.892 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·750/60 = 78.54 rad/s
T = P/ω = 5000/78.54 = 63.66 N·m

Polar second moment of area

J = π(50⁴ − 38⁴)/32 = 408885 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 63662·25/408885 = 3.892 MPa

Answer

τmax = 3.892 MPa
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 = 82.5 kN, R2 = 62.5 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 = 141, MC = 234, MD = 279 kN·m, ME = 226, MF = 125, MG = 0

Answer

Mark D: MD = 279 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 = 5 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 = 5, MC = 10, MD = 0, ME = −10, MF = −20 kN·m, MG = 0

Answer

Mark F: MF = −20 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 − 2827·72)/(36000 − 2827) = 124.1 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: Ix = 226.8 × 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 = (48000·120 − 2400·84)/(48000 − 2400) = 121.9 mm

Full plate about ȳ

200·240³/12 + 48000·−1.895² = 230.6 × 10⁶ mm⁴

Hole about ȳ

60·40³/12 + 2400·−37.89² = 3.766 × 10⁶ mm⁴

Subtract

Ix = 226.8 × 10⁶ mm⁴

Answer

Ix = 226.8 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τmax = 41.23 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.

σxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((120 − 100)/2)² + −40²) = 41.23 MPa

Answer

τmax = 41.23 MPa
10.Mohr’s circle & stress transformation

Answer: σθ = 108.7 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

σθ = 65 + 25·0.1736 + 40·0.9848 = 108.7 MPa

Answer

σθ = 108.7 MPa
11.Beam deflection

Answer: θA = 0.0031 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 4 N/mm, L = 2500 mm
EI = 8.4 × 10¹¹ N·mm²

End slope

θA = wL³/(24EI) = 4·2500³/(24·8.4 × 10¹¹) = 0.0031 rad

Answer

θA = 0.0031 rad

That is 0.1776°.

12.Beam deflection

Answer: MA = 16.88 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 = 3·15·3/8 = 16.88 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

MA = wL²/2 − RB·L = 67.5 − 50.63 = 16.88 kN·m

Answer

MA = 16.88 kN·m
13.Column buckling

Answer: Pcr = 21.48 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·33333/1750² = 21485 N = 21.48 kN

Answer

Pcr = 21.48 kN
14.Column buckling

Answer: d = 80.72 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 = 3·30000 = 90000 N
KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·4000²/(π³·70000))^(1/4) = 80.72 mm

Answer

d = 80.72 mm

Check: the slenderness KL/r is 198.2, so the column is slender and Euler applies.

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