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 = ?12400 mm300 mmØ20Ø30δ = 0.2 mm

A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: copper, Ø30 mm, 300 mm, E = 120 GPaδ = 0.2 mm
P = kN
2.Axial stress & strainHard
20 kN12600 mm400 mmØ25Ø50σ2​ = ?

A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 400 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 400 mm, E = 110 GPaP = 20 kN
σ2 = MPa
3.Torsion of shaftsMedium
T = 200 N·md = 50 mmL = 2.5 mφ = ?

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 = 50 mmL = 2.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?5 kW, 3000 rpm

A hollow shaft must transmit 5 kW at 3000 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter do.

P = 5 kWN = 3000 rpmdi/do = 0.7τallow = 80 MPa
do = mm
5.Shear force & bending momentMedium
4 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 @ 2 m4 kN/m, 2–5 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 = 5 kN/mP = 20 kNABsection1 m1 m2 m

Find the bending moment Mx at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 20 kN at 1 mw = 5 kN/m from 0 to 4 mx = 2 m
Mx = kN·m
7.Centroid & second moment of areaMedium
xD = 150d = 138dimensions in mm

A circular tube has outside diameter D = 150 mm and inside diameter d = 138 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 150 mmd = 138 mm
r = mm
8.Centroid & second moment of areaHard
2002530010xsagging Mdimensions in mm

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 30 MPa in tension and 120 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.

bf = 200 mmtf = 25 mmtw = 10 mmhw = 300 mmσt,allow = 30 MPaσc,allow = 120 MPa
Mallow = kN·m
9.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 80τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 80 MPa and τₓᵧ = 20 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 110 MPaσy = 80 MPaτxy = 20 MPaσz = 0
τabs = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 30σy​ = 90τxy​ = 40MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 90 MPa and τₓᵧ = 40 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 30 MPaσy = 90 MPaτxy = 40 MPaσY = 250 MPa
n =
11.Beam deflectionMedium
w = 5 kN/mABθ6 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 = 5 kN/mL = 6 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴
θA = rad
12.Beam deflectionHard
w = 3 kN/mABδ5 m

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth hreq of the section.

w = 3 kN/mL = 5 mE = 11 GPa (timber)b = 75 mm
hreq = mm
13.Column bucklingMedium
PL = 3 m40 × 100 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 = 3 mE = 200 GPa (steel)b × h = 40 × 100 mm
Pcr = kN
14.Column bucklingHard
P = 75 kNL = 4 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 75 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 75 kNn = 2.5L = 4 mE = 200 GPa (steel)
d = mm

3Solutions

1.Axial stress & strain

Answer: P = 20.2 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 = 314.2 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(120000·706.9) = 3.537 × 10⁻⁶ mm/N

Load

P = 0.2/9.903 × 10⁻⁶ = 20196 N = 20.2 kN

Answer

P = 20.2 kN
2.Axial stress & strain

Answer: σ2 = 7.817 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 = 200000·490.9/600 = 163625 N/mm
k₂ = 110000·1963/400 = 539961 N/mm

Share of segment 2

F2 = 20000·539961/703586 = 15349 N

Stress

σ2 = 15349/1963 = 7.817 MPa (compression)

Answer

σ2 = 7.817 MPa
3.Torsion of shafts

Answer: φ = 1.796 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2500/(26000·613592) = 0.03134 rad
φ = 0.03134·180/π = 1.796°

Answer

φ = 1.796°
4.Torsion of shafts

Answer: do = 11.01 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·3000/60 = 314.2 rad/s
T = P/ω = 5000/314.2 = 15.92 N·m

Outer diameter from the allowable stress

T = 15.92 N·m = 15915 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·15915/(π·80·0.7599)) = 11.01 mm
di = 0.7·11.01 = 7.705 mm

Answer

do = 11.01 mm

A solid shaft would need d = 10.04 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: C (x = 2 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 = 18.33 kN, R2 = 13.67 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 = 18.33, MC = 36.67 kN·m, MD = 33, ME = 25.33, MF = 13.67, MG = 0

Answer

Mark C: MC = 36.67 kN·m
6.Shear force & bending moment

Answer: Mx = 20 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 = 5·4 = 20 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 20·1 + 5·4·2
RB = 60/4 = 15 kN
ΣFy = 0: RA = 40 − 15 = 25 kN
w = 5 kN/mP = 20 kNABRA​ = 25RB​ = 15section1 m1 m2 m

Cut at x = 2 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 = 25·2 − 20·(2 − 1) − 5·2·1 = 20 kN·m

Answer

Mx = 20 kN·m
7.Centroid & second moment of area

Answer: r = 50.96 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 = 2714 mm²
I = π(D⁴ − d⁴)/64 = 7.048 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(7.048 × 10⁶/2714) = 50.96 mm

Answer

r = 50.96 mm
8.Centroid & second moment of area

Answer: Mallow = 8.619 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 = (5000·312.5 + 3000·150)/8000 = 251.6 mm
top edge: c = 325 − 251.6 = 73.44 mm

Second moment of area

I = 200·25³/12 + 5000·60.94² + 10·300³/12 + 3000·101.6² = 7.227 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 120·7.227 × 10⁷/73.44 = 118.1 kN·m
bottom edge: M = 30·7.227 × 10⁷/251.6 = 8.619 kN·m

Governing value

Mallow = min = 8.619 kN·m (bottom edge)

Answer

Mallow = 8.619 kN·m
9.Mohr’s circle & stress transformation

Answer: τabs = 60 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 = 95 MPa, R = √(15² + 20²) = 25 MPa
σ1 = 120 MPa, σ2 = 70 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(50, 120, 70)/2 = 60 MPa

Answer

τabs = 60 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: n = 2.373

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 = 60 MPa, R = 50 MPa
σ1 = 110 MPa, σ2 = 10 MPa, σ3 = 0

Von Mises

σvM = √(110² − 110·10 + 10²) = 105.4 MPa
n = 250/105.4 = 2.373

Answer

n = 2.373

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

11.Beam deflection

Answer: θA = 0.002679 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 = 5 N/mm, L = 6000 mm
EI = 1.68 × 10¹³ N·mm²

End slope

θA = wL³/(24EI) = 5·6000³/(24·1.68 × 10¹³) = 0.002679 rad

Answer

θA = 0.002679 rad

That is 0.1535°.

12.Beam deflection

Answer: hreq = 294.6 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 = 3 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·3·5000⁴/(384·11000·13.89) = 1.598 × 10⁸ mm⁴

Required depth

bh³/12 = Ireq → h = (12·1.598 × 10⁸/75)1/3 = 294.6 mm

Answer

hreq = 294.6 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: Pcr = 117 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·533333/3000² = 116973 N = 117 kN

Answer

Pcr = 117 kN
14.Column buckling

Answer: d = 74.59 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 = 2.5·75000 = 187500 N
KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·187500·4000²/(π³·200000))^(1/4) = 74.59 mm

Answer

d = 74.59 mm

Check: the slenderness KL/r is 214.5, so the column is slender and Euler applies.

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