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: ε_T = 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

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
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/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
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: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 600 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

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

P = 10 kNsegment 1: Ø25 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 40 mm, inner diameter 28 mm, length 2 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.

T = 200 N·md_o = 40 mmd_i = 28 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1800 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1800 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 8 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 6 kN/m from 0 to 10 mx = 8 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 60 kN at 3 mw = 3 kN/m from 0 to 4 mx = 2.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 180 mm and inside diameter d = 152 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 180 mmd = 152 mm

r = mm

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 40 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 300 mmt_f = 20 mmt_w = 14 mmh_w = 180 mmM = 40 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 50 MPa, σᵧ = 90 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 15° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 50 MPaσ_y = 90 MPaτ_xy = −20 MPaθ = 15°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 25 mm carries an axial compressive force of 140 kN together with a torque of 400 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 25 mmF = −140 kNT = 400 N·m

σ₁ = MPa

11.Beam deflectionMedium

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

w = 10 kN/mL = 1.5 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the maximum deflection δ_max of the beam.

P = 23 kNb = 1.5 mL = 5 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 4 mE = 70 GPa (aluminium)b × h = 40 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 2 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 65.75 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, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·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 = 1257 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·2827) = 1.415 × 10⁻⁶ mm/N

Load

P = 0.25/3.802 × 10⁻⁶ = 65754 N = 65.75 kN

Answer

P = 65.75 kN

2.Axial stress & strain

Answer: σ_max = 88.42 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 = 20.37 MPa

σ₂ = 88.42 MPa

Largest stress

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

σ_max = 88.42 MPa

Answer

σ_max = 88.42 MPa

3.Torsion of shafts

Answer: φ = 1.5 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/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 − 28⁴)/32 = 190984 mm⁴

Angle of twist

T = 200 N·m = 200000 N·mm

L = 2 m = 2000 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 200000·2000/(80000·190984) = 0.02618 rad

φ = 0.02618·180/π = 1.5°

Answer

φ = 1.5°

4.Torsion of shafts

Answer: d = 40.73 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 100000/188.5 = 530.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·530516/(π·40)) = 40.73 mm

Answer

d = 40.73 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_x = 48 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 = 6·10 = 60 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 6·10·5

R_B = 300/10 = 30 kN

ΣF_y = 0: R_A = 60 − 30 = 30 kN

Cut at x = 8 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 30·8 − 6·8·4 = 48 kN·m

Answer

M_x = 48 kN·m

6.Shear force & bending moment

Answer: M_x = 43.13 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·4 = 12 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 60·3 + 3·4·2

R_B = 204/4 = 51 kN

ΣF_y = 0: R_A = 72 − 51 = 21 kN

Cut at x = 2.5 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 21·2.5 − 3·2.5·1.25 = 43.13 kN·m

Answer

M_x = 43.13 kN·m

7.Centroid & second moment of area

Answer: r = 58.9 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 = I_c + 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 = 7301 mm²

I = π(D⁴ − d⁴)/64 = 2.533 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(2.533 × 10⁷/7301) = 58.9 mm

Answer

r = 58.9 mm

8.Centroid & second moment of area

Answer: σ_top = 63.96 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: A_f = 300·20 = 6000 mm², y_f = 180 + 20/2 = 190 mm

web: A_w = 14·180 = 2520 mm², y_w = 180/2 = 90 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·190 + 2520·90)/8520 = 160.4 mm

Parallel-axis theorem

flange: 300·20³/12 + 6000·(190 − 160.4)² = 5.449 × 10⁶ mm⁴

web: 14·180³/12 + 2520·(90 − 160.4)² = 1.93 × 10⁷ mm⁴

I_x = 5.449 × 10⁶ + 1.93 × 10⁷ = 2.475 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 200 − 160.4 = 39.58 mm

σ = M·y/I = 4 × 10⁷·39.58/2.475 × 10⁷ = 63.96 MPa

Answer

σ_top = 63.96 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 97.32 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 for the y′ face

σ_y′ = 70 − (−20)·0.866 − (−20)·0.5 = 97.32 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 42.68 MPa, σ_x′ + σ_y′ = 140 = σ_x + σ_y

Answer

σ_y′ = 97.32 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 50.62 MPa

Stresses at the surface

σ = F/A = −140000/(π·25²/4) = −140000/490.9 = −285.2 MPa

τ = 16T/(πd³) = 16·400000/(π·25³) = 130.4 MPa

Mohr’s circle

C = σ/2 = −142.6 MPa

R = √((−142.6)² + 130.4²) = 193.2 MPa

σ₁ = C + R = 50.62 MPa

σ₂ = C − R = −335.8 MPa

Answer

σ₁ = 50.62 MPa

11.Beam deflection

Answer: δ_max = 1.665 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 = 10 N/mm, L = 1500 mm

EI = 3.96 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·1500⁴/(384·3.96 × 10¹¹) = 1.665 mm

Answer

δ_max = 1.665 mm

12.Beam deflection

Answer: δ_max = 51.98 mm

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 = 23000 N, b = 1500 mm, L = 5000 mm

EI = 9.24 × 10¹¹ N·mm²

Position of the maximum

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

x = √((L² − b²)/3) = √((5000² − 1500²)/3) = 2754 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 51.98 mm

Answer

δ_max = 51.98 mm

13.Column buckling

Answer: P_cr = 73.69 kN

Euler buckling: a slender column becomes unstable at P_cr = 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.

I_min = h·b³/12 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·426667/2000² = 73693 N = 73.69 kN

Answer

P_cr = 73.69 kN

14.Column buckling

Answer: σ_cr = 69.09 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 1·2000 = 2000 mm

KL/r = 2000/20 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

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