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
P = ?12600 mm500 mmØ25Ø30δ = 0.4 mm

A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaδ = 0.4 mm
P = kN
2.Axial stress & strainHard
Ø 25 mmBrass2 mØ 12 mmSteel1 mP = 25 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 = 25 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
3.Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 2.5 mφ = ?

A solid steel 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 = 80 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?15 kW, 750 rpm

A hollow shaft must transmit 15 kW at 750 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.

P = 15 kWN = 750 rpmdi/do = 0.6τallow = 75 MPa
do = mm
5.Shear force & bending momentMedium
35 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.

35 kN @ 3 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)
6.Shear force & bending momentHard
w = 6 kN/mP = 50 kNAB1 m9 m

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.

L = 10 mP = 50 kN at 1 mw = 6 kN/m from 0 to 10 m
Mmax = kN·m
7.Centroid & second moment of areaMedium
220102608xdimensions in mm

Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.

bf = 220 mmtf = 10 mmtw = 8 mmhw = 260 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
1201514012xsagging Mdimensions in mm

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 40 MPa in tension and 150 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.

bf = 120 mmtf = 15 mmtw = 12 mmhw = 140 mmσt,allow = 40 MPaσc,allow = 150 MPa
Mallow = kN·m
9.Mohr’s circle & stress transformationMedium
σx​ = 40σy​ = 10τxy​ = 20MPa

A point in a loaded part has σₓ = 40 MPa, σᵧ = 10 MPa and τₓᵧ = 20 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 40 MPaσy = 10 MPaτxy = 20 MPa
θp = °
10.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = −70τxy​ = −30MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 70 MPaσy = −70 MPaτxy = −30 MPaσY = 460 MPa
n =
11.Beam deflectionMedium
P = 10 kNABδ1 m1.5 m

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δB at the free end B.

P = 10 kNa = 1 mL = 2.5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
δB = mm
12.Beam deflectionHard
w = 6 kN/mδL = 2.5 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 = 6 kN/mL = 2.5 mE = 11 GPa (timber)
Ireq = × 10⁶ mm⁴
13.Column bucklingMedium
PL = 1.5 md = 25 mmcross-section

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

L = 1.5 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
14.Column bucklingHard
P = 150 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 150 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 150 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm

3Solutions

1.Axial stress & strain

Answer: P = 24.67 kN

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

A₁ = 490.9 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·706.9) = 1.011 × 10⁻⁵ mm/N

Load

P = 0.4/1.622 × 10⁻⁵ = 24666 N = 24.67 kN

Answer

P = 24.67 kN
2.Axial stress & strain

Answer: σmax = 221 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·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 50.93 MPa
σ2 = 221 MPa

Largest stress

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

σmax = 221 MPa

Answer

σmax = 221 MPa
3.Torsion of shafts

Answer: φ = 5.343 °

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/(80000·251327) = 0.09325 rad
φ = 0.09325·180/π = 5.343°

Answer

φ = 5.343°
4.Torsion of shafts

Answer: do = 24.61 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 = 15000/78.54 = 191 N·m

Outer diameter from the allowable stress

T = 191 N·m = 190986 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·190986/(π·75·0.8704)) = 24.61 mm
di = 0.6·24.61 = 14.76 mm

Answer

do = 24.61 mm

A solid shaft would need d = 23.49 mm; the hollow one is a little larger but much lighter.

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 = 27.5 kN, R2 = 22.5 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 = 27.5, MC = 55, MD = 67.5 kN·m, ME = 45, MF = 22.5, MG = 0

Answer

Mark D: MD = 67.5 kN·m
6.Shear force & bending moment

Answer: Mmax = 102.1 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·10 = 60 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 50·1 + 6·10·5
RB = 350/10 = 35 kN
ΣFy = 0: RA = 110 − 35 = 75 kN

Where is V = 0?

RA − P − w·x0 = 0 → x0 = (75 − 50)/6 = 4.167 m

Maximum moment

Mmax = RA·x0 − P·(x0 − 1) − w·x0²/2 = 102.1 kN·m

Answer

Mmax = 102.1 kN·m

It occurs at x = 4.167 m from A.

7.Centroid & second moment of area

Answer: Ix = 31.22 × 10⁶ mm⁴

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 = 220·10 = 2200 mm², yf = 260 + 10/2 = 265 mm
web: Aw = 8·260 = 2080 mm², yw = 260/2 = 130 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (2200·265 + 2080·130)/4280 = 199.4 mm

Parallel-axis theorem

flange: 220·10³/12 + 2200·(265 − 199.4)² = 9.488 × 10⁶ mm⁴
web: 8·260³/12 + 2080·(130 − 199.4)² = 2.173 × 10⁷ mm⁴
Ix = 9.488 × 10⁶ + 2.173 × 10⁷ = 3.122 × 10⁷ mm⁴ = 31.22 × 10⁶ mm⁴

Answer

Ix = 31.22 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: Mallow = 2.906 kN·m

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.

Centroid

ȳ = (1800·147.5 + 1680·70)/3480 = 110.1 mm
top edge: c = 155 − 110.1 = 44.91 mm

Second moment of area

I = 120·15³/12 + 1800·37.41² + 12·140³/12 + 1680·40.09² = 7.997 × 10⁶ mm⁴

Moment allowed by each edge

Sagging: the top is in compression, the bottom in tension.

top edge: M = 150·7.997 × 10⁶/44.91 = 26.71 kN·m
bottom edge: M = 40·7.997 × 10⁶/110.1 = 2.906 kN·m

Governing value

Mallow = min = 2.906 kN·m (bottom edge)

Answer

Mallow = 2.906 kN·m
9.Mohr’s circle & stress transformation

Answer: θp = 26.57 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·20, 40 − 10) = 53.13°

Angle on the element

θp = 26.57°

Answer

θp = 26.57°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 3.02

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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

στxyCσ1​σ2​

Principal stresses

C = 0 MPa, R = 76.16 MPa
σ1 = 76.16 MPa, σ2 = −76.16 MPa, σ3 = 0

Tresca

σmax − σmin = 152.3 MPa
n = 460/152.3 = 3.02

Answer

n = 3.02

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δB = 6.373 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 = 10000 N, a = 1000 mm, L = 2500 mm
EI = 1.7 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δP = Pa³/(3EI) = 1.961 mm
θP = Pa²/(2EI) = 0.002941 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 6.373 mm

Answer

δB = 6.373 mm
12.Beam deflection

Answer: Ireq = 39.95 × 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 = 2500/360 = 6.944 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·6·2500⁴/(384·11000·6.944) = 3.995 × 10⁷ mm⁴
Ireq = 39.95 × 10⁶ mm⁴

Answer

Ireq = 39.95 × 10⁶ mm⁴
13.Column buckling

Answer: Pcr = 16.82 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·19175/1500² = 16822 N = 16.82 kN

Answer

Pcr = 16.82 kN
14.Column buckling

Answer: d = 54.32 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 = 2.5·150000 = 375000 N
KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·375000·1500²/(π³·200000))^(1/4) = 54.32 mm

Answer

d = 54.32 mm

Check: the slenderness KL/r is 110.5, so the column is slender and Euler applies.

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