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 = ?12500 mm400 mmØ25Ø60δ = 0.25 mm

A bar is made of a steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaδ = 0.25 mm
P = kN
2.Axial stress & strainHard
Ø 40 mmBrass1 mØ 12 mmBrass1.5 mP = 115 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 115 kNsegment 1: Ø40 mm, 1 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1.5 m, brass (E = 100 GPa)
δ = mm
3.Torsion of shaftsMedium
T = 100 N·md = 30 mmL = 0.5 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 = 100 N·md = 30 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = 50 mmT0​ = 750 N·mACBa = 2 mb = 1 m

A solid brass shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 750 N·m is applied at C, 2 m from A and 1 m from B. Find the maximum shear stress τmax in the shaft.

d = 50 mmT0 = 750 N·ma = 2 mb = 1 m
τmax = MPa
5.Shear force & bending momentMedium
10 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 @ 2 m55 kN @ 4 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
12 kN/m10 kN10 kN20 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.

10 kN @ 0 m10 kN @ 6 m20 kN @ 3 m12 kN/m, 1–4 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)
7.Centroid & second moment of areaMedium
120 mm240 mm144 mmØ50xIx​ = ?

A 120 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 144 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 120 mmh = 240 mmd = 50 mmhole centre: y = 144 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
2801528016xM = 20 kN·mdimensions in mm

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 20 kN·m. Find the bending stress σbottom at the bottom edge.

bf = 280 mmtf = 15 mmtw = 16 mmhw = 280 mmM = 20 kN·m
σbottom = MPa
9.Mohr’s circle & stress transformationMedium
σx​ = 10σy​ = 70MPaτxy​ = ?

A point has σₓ = 10 MPa and σᵧ = 70 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 130 MPa?

σx = 10 MPaσy = 70 MPaσ1 = 130 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
35°D = 400 mm, t = 12 mm, p = 1.5 MPap

A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 12 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.

D = 400 mmt = 12 mmp = 1.5 MPaα = 35°
σn = MPa
11.Beam deflectionMedium
w = 15 kN/mABθ4 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 = 15 kN/mL = 4 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
θA = rad
12.Beam deflectionHard
w = 5 kN/mAB8 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 reaction RB at the prop.

L = 8 mw = 5 kN/m
RB = kN
13.Column bucklingMedium
PL = 2.5 md = 80 mmcross-section

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

L = 2.5 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
14.Column bucklingHard
PL = 2 md = 30 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 2 mE = 200 GPa (steel)d = 30 mm
σcr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 38.42 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 = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·490.9) = 5.093 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(100000·2827) = 1.415 × 10⁻⁶ mm/N

Load

P = 0.25/6.508 × 10⁻⁶ = 38416 N = 38.42 kN

Answer

P = 38.42 kN
2.Axial stress & strain

Answer: δ = 16.17 mm

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·12²/4 = 113.1 mm²
σ1 = 91.51 MPa
σ2 = 1017 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 91.51·1000/100000 = 0.9151 mm
δ2 = σ2L2/E2 = 1017·1500/100000 = 15.25 mm
δ = δ1 + δ2 = 16.17 mm

Answer

δ = 16.17 mm
3.Torsion of shafts

Answer: φ = 0.9237 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·500/(39000·79522) = 0.01612 rad
φ = 0.01612·180/π = 0.9237°

Answer

φ = 0.9237°
4.Torsion of shafts

Answer: τmax = 20.37 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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 750
TA·a = TB·b
TA = 750·1/3 = 250 N·m
TB = 750·2/3 = 500 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τmax = 16·500000/(π·50³) = 20.37 MPa

Answer

τmax = 20.37 MPa
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 = 45 kN, R2 = 20 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 = 90, MC = 160 kN·m, MD = 120, ME = 80, MF = 40, MG = 0

Answer

Mark C: MC = 160 kN·m
6.Shear force & bending moment

Answer: E (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 = 31.33 kN, R2 = 44.67 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 = −10, MC = 5.333, MD = 8.667, ME = −20 kN·m, MF = −10, MG = 0

Answer

Mark E: ME = −20 kN·m
7.Centroid & second moment of area

Answer: Ix = 136.7 × 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 − 1963·144)/(28800 − 1963) = 118.2 mm

Full plate about ȳ

120·240³/12 + 28800·1.756² = 138.3 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·25.76² = 1.609 × 10⁶ mm⁴

Subtract

Ix = 136.7 × 10⁶ mm⁴

Answer

Ix = 136.7 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: σbottom = 55.25 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.

Split into two rectangles

Measure every y from the bottom edge.

flange: Af = 280·15 = 4200 mm², yf = 280 + 15/2 = 287.5 mm
web: Aw = 16·280 = 4480 mm², yw = 280/2 = 140 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (4200·287.5 + 4480·140)/8680 = 211.4 mm

Parallel-axis theorem

flange: 280·15³/12 + 4200·(287.5 − 211.4)² = 2.442 × 10⁷ mm⁴
web: 16·280³/12 + 4480·(140 − 211.4)² = 5.209 × 10⁷ mm⁴
Ix = 2.442 × 10⁷ + 5.209 × 10⁷ = 7.651 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 211.4 mm
σ = M·y/I = 2 × 10⁷·211.4/7.651 × 10⁷ = 55.25 MPa

Answer

σbottom = 55.25 MPa
9.Mohr’s circle & stress transformation

Answer: τxy = 84.85 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 = (10 + 70)/2 = 40 MPa
R = σ1 − C = 130 − 40 = 90 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(90² − 30²) = 84.85 MPa

Answer

τxy = 84.85 MPa
10.Mohr’s circle & stress transformation

Answer: σn = 20.89 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 = 400/2 = 200 mm
σh = pr/t = 1.5·200/12 = 25 MPa
σa = pr/(2t) = 12.5 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.

σn = 12.5·sin²35° + 25·cos²35° = 20.89 MPa

Answer

σn = 20.89 MPa
11.Beam deflection

Answer: θA = 0.004762 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 = 15 N/mm, L = 4000 mm
EI = 8.4 × 10¹² N·mm²

End slope

θA = wL³/(24EI) = 15·4000³/(24·8.4 × 10¹²) = 0.004762 rad

Answer

θA = 0.004762 rad

That is 0.2728°.

12.Beam deflection

Answer: RB = 15 kN

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 = 3·5·8/8 = 15 kN

Answer

RB = 15 kN
13.Column buckling

Answer: Pcr = 222.3 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 round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2500² = 222253 N = 222.3 kN

Answer

Pcr = 222.3 kN
14.Column buckling

Answer: σcr = 6.94 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 2·2000 = 4000 mm
KL/r = 4000/7.5 = 533.3

Critical stress

σcr = π²E/(KL/r)² = π²·200000/533.3² = 6.94 MPa

Answer

σcr = 6.94 MPa

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