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
Ø 20 mmCopper2 mP = 110 kN

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 110 kNd = 20 mmL = 2 mE = 117 GPa
ε = µε
2.Axial stress & strainHard
P = ?Ø20100 GPaØ120/50300 mm

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø20 mm, E = 200 GPa, σallow = 200 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 100 MPa
Pmax = kN
3.Torsion of shaftsMedium
T = 200 N·md = 25 mmL = 3 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 = 200 N·md = 25 mmL = 3 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = 30 mmTB​ = 100 N·mTC​ = 250 N·mABC3 m0.5 m

A solid steel shaft of diameter 30 mm is fixed at A. A torque of 100 N·m acts at B, 3 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the magnitude of the angle of twist φC of end C, in degrees.

d = 30 mmTB = 100 N·mTC = 250 N·mAB = 3 mBC = 0.5 mG = 80 GPa
φC = °
5.Shear force & bending momentMedium
P1​ = 65 kNP2​ = 55 kNAB1.5 m0.5 m3 m

The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.

L = 5 mP1 = 65 kN at 1.5 mP2 = 55 kN at 2 m
Mmax = kN·m
6.Shear force & bending momentHard
w = 10 kN/mP = 10 kNABsection1.5 m2.5 m

Find the bending moment Mx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 10 kN at 1.5 mw = 10 kN/m from 0 to 4 mx = 1.5 m
Mx = kN·m
7.Centroid & second moment of areaMedium
2504202014xdimensions in mm

A symmetric I-section has flanges 250 mm wide and 20 mm thick, an overall depth of 420 mm and a web 14 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.

B = 250 mmH = 420 mmtf = 20 mmtw = 14 mm
Ix = × 10⁶ mm⁴
8.Centroid & second moment of areaHard
130250243902216xdimensions in mm

A welded I-section has a top flange 130 × 24 mm, a web 390 mm high and 16 mm thick, and a bottom flange 250 × 22 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 130 × 24 mmweb: 16 × 390 mmbottom flange: 250 × 22 mm
Ix = × 10⁶ mm⁴
9.Mohr’s circle & stress transformationMedium
σx​ = 20MPaτxy​ = ?

A point has σₓ = 20 MPa and σᵧ = 0 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 75 MPa?

σx = 20 MPaσy = 0 MPaσ1 = 75 MPa
τxy = MPa
10.Mohr’s circle & stress transformationHard
σx​ = 150σy​ = 30τxy​ = 30MPa

A point in a loaded part has σₓ = 150 MPa, σᵧ = 30 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σx = 150 MPaσy = 30 MPaτxy = 30 MPaθ = 35°
σθ = MPa
11.Beam deflectionMedium
P = 22 kNABδ1 m1 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 = 22 kNa = 1 mL = 2 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴
δB = mm
12.Beam deflectionHard
P = 6 kNABδ3.5 m2.5 m

A simply supported beam carries a point load P at b = 2.5 m from support B. Find the maximum deflection δmax of the beam.

P = 6 kNb = 2.5 mL = 6 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴
δmax = mm
13.Column bucklingMedium
PL = 1.5 m20 × 50 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)b × h = 20 × 50 mm
Pcr = kN
14.Column bucklingHard
PL = 2.5 md = 50 mmcross-section

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

L = 2.5 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 2993 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·20²/4 = 314.2 mm²
σ = P/A = 110·1000/314.2 = 350.1 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 350.1/117000 = 0.002993 = 2993 µε

Answer

ε = 2993 µε
2.Axial stress & strain

Answer: Pmax = 997.5 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

Ar = π·20²/4 = 314.2 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·9346 = 9.975 × 10⁸ N

Governing strain

ε = min(0.001, 0.001) = 0.001 (brass)

Largest load

Pmax = 0.001·9.975 × 10⁸ = 997456 N = 997.5 kN

Answer

Pmax = 997.5 kN
3.Torsion of shafts

Answer: φ = 22.99 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·3000/(39000·38350) = 0.4012 rad
φ = 0.4012·180/π = 22.99°

Answer

φ = 22.99°
4.Torsion of shafts

Answer: φC = 10.58 °

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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 250 N·m
TAB = TC + TB = 250 + 100 = 350 N·m

Twist of each segment

J = π·30⁴/32 = 79522 mm⁴
G = 80 GPa = 80000 MPa
φAB = 350000·3000/(80000·79522) = 0.165 rad
φBC = 250000·500/(80000·79522) = 0.01965 rad

Twist of end C

φC = 0.165 + 0.01965 = 0.1847 rad → |φC| = 10.58°

Answer

φC = 10.58°
5.Shear force & bending moment

Answer: Mmax = 124.5 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.

ΣMA = 0: RB·5 = 65·1.5 + 55·2
RB = 207.5/5 = 41.5 kN
ΣFy = 0: RA = 120 − 41.5 = 78.5 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·1.5 = 78.5·1.5 = 117.8 kN·m
under P2: M = RB·3 = 41.5·3 = 124.5 kN·m

Largest value

Mmax = 124.5 kN·m (under P2)

Answer

Mmax = 124.5 kN·m
6.Shear force & bending moment

Answer: Mx = 28.13 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 = 10·4 = 40 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 10·1.5 + 10·4·2
RB = 95/4 = 23.75 kN
ΣFy = 0: RA = 50 − 23.75 = 26.25 kN
w = 10 kN/mP = 10 kNABRA​ = 26.25RB​ = 23.75section1.5 m2.5 m

Cut at x = 1.5 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 = 26.25·1.5 − 10·1.5·0.75 = 28.13 kN·m

Answer

Mx = 28.13 kN·m
7.Centroid & second moment of area

Answer: Ix = 464.4 × 10⁶ mm⁴

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.

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 = 420 − 2·20 = 380 mm
Ix = (B·H³ − (B − tw)·hw³)/12 = (250·420³ − 236·380³)/12
Ix = 4.644 × 10⁸ mm⁴ = 464.4 × 10⁶ mm⁴

Answer

Ix = 464.4 × 10⁶ mm⁴
8.Centroid & second moment of area

Answer: Ix = 430.6 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 3120 mm², y = 424 mm
web: A = 6240 mm², y = 217 mm
bottom flange: A = 5500 mm², y = 11 mm

Centroid

ȳ = (3120·424 + 6240·217 + 5500·11)/14860 = 184.2 mm

Parallel-axis theorem

top flange: 130·24³/12 + 3120·239.8² = 1.795 × 10⁸ mm⁴
web: 16·390³/12 + 6240·32.78² = 8.58 × 10⁷ mm⁴
bottom flange: 250·22³/12 + 5500·173.2² = 1.652 × 10⁸ mm⁴
Ix = 1.795 × 10⁸ + 8.58 × 10⁷ + 1.652 × 10⁸
Ix = 4.306 × 10⁸ mm⁴ = 430.6 × 10⁶ mm⁴

Answer

Ix = 430.6 × 10⁶ mm⁴
9.Mohr’s circle & stress transformation

Answer: τxy = 64.23 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.

Required radius

C = (20 + 0)/2 = 10 MPa
R = σ1 − C = 75 − 10 = 65 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 10²) = 64.23 MPa

Answer

τxy = 64.23 MPa
10.Mohr’s circle & stress transformation

Answer: σθ = 138.7 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 = 90 + 60·0.342 + 30·0.9397 = 138.7 MPa

Answer

σθ = 138.7 MPa
11.Beam deflection

Answer: δB = 2.546 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 = 22000 N, a = 1000 mm, L = 2000 mm
EI = 7.2 × 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) = 1.019 mm
θP = Pa²/(2EI) = 0.001528 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 2.546 mm

Answer

δB = 2.546 mm
12.Beam deflection

Answer: δmax = 10.33 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 = 6000 N, b = 2500 mm, L = 6000 mm
EI = 2.52 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((6000² − 2500²)/3) = 3149 mm

Maximum deflection

δmax = Pb(L² − b²)3/2/(9√3·EIL) = 10.33 mm

Answer

δmax = 10.33 mm
13.Column buckling

Answer: Pcr = 29.24 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 = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·33333/1500² = 29243 N = 29.24 kN

Answer

Pcr = 29.24 kN
14.Column buckling

Answer: σcr = 4.318 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 = 2·2500 = 5000 mm
KL/r = 5000/12.5 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σcr = 4.318 MPa

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