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
150 kNØ30120 GPaØ120/70300 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 300 mm
δ = mm
2.Axial stress & strainHard
Ø 40 mmSteel1.5 mØ 16 mmBrass1.5 mP = 105 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 105 kNsegment 1: Ø40 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1.5 m, brass (E = 100 GPa)
σmax = MPa
3.Torsion of shaftsMedium
do​ = ?T = 750 N·m

A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

T = 750 N·mdi/do = 0.7τallow = 40 MPa
do = mm
4.Torsion of shaftsHard
d = ?10 kW, 750 rpm

A solid brass shaft transmits 10 kW at 750 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.

P = 10 kWN = 750 rpmτallow = 50 MPaθallow = 0.25°/mG = 39 GPa
d = mm
5.Shear force & bending momentMedium
25 kN10 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 @ 4 m10 kN @ 8 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
10 kN60 kN55 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.

10 kN @ 0 m60 kN @ 12 m55 kN @ 6 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)
7.Centroid & second moment of areaMedium
120 mm240 mm168 mmØ60xIx​ = ?

A 120 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 = 120 mmh = 240 mmd = 60 mmhole centre: y = 168 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
1302202212xM = 60 kN·mdimensions in mm

A symmetric I-section has flanges 130 mm wide and 22 mm thick, an overall depth of 220 mm and a web 12 mm thick. It carries a bending moment of 60 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.

B = 130 mmH = 220 mmtf = 22 mmtw = 12 mmM = 60 kN·m
σj = MPa
9.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 90τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 90 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 110 MPaσy = 90 MPaτxy = 30 MPaσz = 0
τabs = MPa
10.Mohr’s circle & stress transformationHard
30°D = 600 mm, t = 6 mm, p = 1.5 MPap

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 6 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.

D = 600 mmt = 6 mmp = 1.5 MPaα = 30°
|τn| = MPa
11.Beam deflectionMedium
w = 8 kN/mδL = 3 m

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

w = 8 kN/mL = 3 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴
δmax = mm
12.Beam deflectionHard
w = 5 kN/mδL = 4 m

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 = 5 kN/mL = 4 mE = 200 GPa (steel)
Ireq = × 10⁶ mm⁴
13.Column bucklingMedium
PL = 2 m30 × 80 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 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
14.Column bucklingHard
PL = 3 md = 50 mmcross-section

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 = 3 mE = 200 GPa (steel)d = 50 mm
σcr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.04341 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·30²/4 = 706.9 mm²
At = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Shortening

δ = 150000·300/1.037 × 10⁹ = 0.04341 mm

Answer

δ = 0.04341 mm
2.Axial stress & strain

Answer: σmax = 522.2 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 83.56 MPa
σ2 = 522.2 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 522.2 MPa

Answer

σmax = 522.2 MPa
3.Torsion of shafts

Answer: do = 50.09 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.

Outer diameter from the allowable stress

T = 750 N·m = 750000 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·750000/(π·40·0.7599)) = 50.09 mm
di = 0.7·50.09 = 35.06 mm

Answer

do = 50.09 mm

A solid shaft would need d = 45.71 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 52.54 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 = 10000/78.54 = 127.3 N·m

Diameter for strength

T = 127.3 N·m = 127324 N·mm
dτ = ∛(16·127324/(π·50)) = 23.49 mm

Diameter for stiffness

θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·127324/(π·39000·4.363 × 10⁻⁶))1/4 = 52.54 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(23.49, 52.54) = 52.54 mm

Answer

d = 52.54 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: C (x = 4 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 = 20 kN, R2 = 15 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 = 40, MC = 80 kN·m, MD = 70, ME = 60, MF = 30, MG = 0

Answer

Mark C: MC = 80 kN·m
6.Shear force & bending moment

Answer: F (x = 10 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 = 25 kN, R2 = 100 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, MD = 40, ME = −40, MF = −120 kN·m, MG = 0

Answer

Mark F: MF = −120 kN·m
7.Centroid & second moment of area

Answer: Ix = 130.4 × 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 = (28800·120 − 2827·168)/(28800 − 2827) = 114.8 mm

Full plate about ȳ

120·240³/12 + 28800·5.225² = 139 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·53.23² = 8.646 × 10⁶ mm⁴

Subtract

Ix = 130.4 × 10⁶ mm⁴

Answer

Ix = 130.4 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: σj = 85.51 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.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.

hw = 220 − 2·22 = 176 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (130·220³ − 118·176³)/12
Ix = 6.174 × 10⁷ mm⁴

Bending stress

y = 110 − 22 = 88 mm
σ = M·y/I = 6 × 10⁷·88/6.174 × 10⁷ = 85.51 MPa

Answer

σj = 85.51 MPa
9.Mohr’s circle & stress transformation

Answer: τabs = 65.81 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.

σxyCσ1​σ2​

In-plane principal stresses

C = 100 MPa, R = √(10² + 30²) = 31.62 MPa
σ1 = 131.6 MPa, σ2 = 68.38 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(63.25, 131.6, 68.38)/2 = 65.81 MPa

Answer

τabs = 65.81 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: |τn| = 16.24 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 = 600/2 = 300 mm
σh = pr/t = 1.5·300/6 = 75 MPa
σa = pr/(2t) = 37.5 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.

|τn| = (75 − 37.5)·sin 30°·cos 30° = 16.24 MPa

Answer

|τn| = 16.24 MPa
11.Beam deflection

Answer: δmax = 13.46 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 = 8 N/mm, L = 3000 mm
EI = 6.27 × 10¹¹ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·8·3000⁴/(384·6.27 × 10¹¹) = 13.46 mm

Answer

δmax = 13.46 mm
12.Beam deflection

Answer: Ireq = 7.5 × 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 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·5·4000⁴/(384·200000·11.11) = 7.5 × 10⁶ mm⁴
Ireq = 7.5 × 10⁶ mm⁴

Answer

Ireq = 7.5 × 10⁶ mm⁴
13.Column buckling

Answer: Pcr = 181.3 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·180000/1400² = 181278 N = 181.3 kN

Answer

Pcr = 181.3 kN
14.Column buckling

Answer: σcr = 69.94 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 = 0.7·3000 = 2100 mm
KL/r = 2100/12.5 = 168

Critical stress

σcr = π²E/(KL/r)² = π²·200000/168² = 69.94 MPa

Answer

σcr = 69.94 MPa

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