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 20 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 300 mm, E = 70 GPaδ = 0.2 mm

P = kN

2.Axial stress & strainHard

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

P = 25 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 25 mm and length 2.5 m; segment BC has diameter 20 mm and length 1.2 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 250 N·md₁ = 25 mm, L₁ = 2.5 md₂ = 20 mm, L₂ = 1.2 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 1500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 1500 rpmd_o = 50 mmd_i = 30 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 35 kN·m at x = 3.5 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 35 kN·m clockwise, at 3.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 35 kN at 3 mw = 8 kN/m from 0 to 8 mx = 7.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 84 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmd = 50 mmhole centre: y = 84 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 300 mm plate has a rectangular cut-out 80 mm wide and 50 mm high, centred on the vertical centre line with its centre 105 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 300 mmcut-out: 80 × 50 mmhole centre: y = 105 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −70 MPa, σᵧ = 100 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = −70 MPaσ_y = 100 MPaτ_xy = −30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 70 MPa, σᵧ = 0 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.

σ_x = 70 MPaσ_y = 0 MPaτ_xy = 30 MPaθ = 45°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1.5 m from A. Find the deflection δ_B at the free end B.

P = 25 kNa = 1.5 mL = 3 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 3 mP = 50 kN at 1.5 m

M_A = kN·m

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)d = 30 mm

P_cr = kN

14.Column bucklingHard

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

L = 4 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 31.79 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 = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·314.2) = 4.775 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(70000·2827) = 1.516 × 10⁻⁶ mm/N

Load

P = 0.2/6.29 × 10⁻⁶ = 31794 N = 31.79 kN

Answer

P = 31.79 kN

2.Axial stress & strain

Answer: δ = 1.487 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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 50.93 MPa

σ₂ = 221 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 50.93·1500/200000 = 0.382 mm

δ₂ = σ₂L₂/E₂ = 221·1000/200000 = 1.105 mm

δ = δ₁ + δ₂ = 1.487 mm

Answer

δ = 1.487 mm

3.Torsion of shafts

Answer: τ_max = 159.2 MPa

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.

Stress in each segment

T = 250 N·m = 250000 N·mm

τ_AB = 16·250000/(π·25³) = 81.49 MPa

τ_BC = 16·250000/(π·20³) = 159.2 MPa

Answer

τ_max = 159.2 MPa

4.Torsion of shafts

Answer: τ_max = 22.35 MPa

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 75000/157.1 = 477.5 N·m

Polar second moment of area

J = π(50⁴ − 30⁴)/32 = 534071 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 477465·25/534071 = 22.35 MPa

Answer

τ_max = 22.35 MPa

5.Shear force & bending moment

Answer: M_x = 14.58 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 35/6 = 5.833 kN (upward)

R_A = −R_B = −5.833 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −5.833·3.5 + 35 = 14.58 kN·m

Answer

M_x = 14.58 kN·m

6.Shear force & bending moment

Answer: M_x = 21.56 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·8 = 64 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 35·3 + 8·8·4

R_B = 361/8 = 45.13 kN

ΣF_y = 0: R_A = 99 − 45.13 = 53.88 kN

Cut at x = 7.5 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 = 53.88·7.5 − 35·(7.5 − 3) − 8·7.5·3.75 = 21.56 kN·m

Answer

M_x = 21.56 kN·m

7.Centroid & second moment of area

Answer: I_x = 227.4 × 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 = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (48000·120 − 1963·84)/(48000 − 1963) = 121.5 mm

Full plate about ȳ

200·240³/12 + 48000·−1.535² = 230.5 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·−37.54² = 3.073 × 10⁶ mm⁴

Subtract

I_x = 227.4 × 10⁶ mm⁴

Answer

I_x = 227.4 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 327.8 × 10⁶ mm⁴

Centroid

ȳ = (45000·150 − 4000·105)/(45000 − 4000) = 154.4 mm

Full plate about ȳ

150·300³/12 + 45000·−4.39² = 338.4 × 10⁶ mm⁴

Hole about ȳ

80·50³/12 + 4000·−49.39² = 10.59 × 10⁶ mm⁴

Subtract

I_x = 327.8 × 10⁶ mm⁴

Answer

I_x = 327.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = −80.28 °

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.

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.

2θ_p = atan2(2·−30, −70 − 100) = −160.6°

Angle on the element

θ_p = −80.28°

Answer

θ_p = −80.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 65 MPa

Transformation equation

σ_θ = 35 + 35·0 + 30·1 = 65 MPa

Answer

σ_θ = 65 MPa

11.Beam deflection

Answer: δ_B = 4.185 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 = 25000 N, a = 1500 mm, L = 3000 mm

EI = 1.68 × 10¹³ N·mm²

Deflection at the free end

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

δ_P = Pa³/(3EI) = 1.674 mm

θ_P = Pa²/(2EI) = 0.001674 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 4.185 mm

Answer

δ_B = 4.185 mm

12.Beam deflection

Answer: M_A = 28.13 kN·m

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = 5PL³/(48EI) → R_B = 5P/16 = 5·50/16 = 15.63 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = P·L/2 − R_B·L = 75 − 46.88 = 28.13 kN·m

Answer

M_A = 28.13 kN·m

13.Column buckling

Answer: P_cr = 6.867 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 round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·39761/2000² = 6867 N = 6.867 kN

Answer

P_cr = 6.867 kN

14.Column buckling

Answer: σ_cr = 6.94 MPa

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 = 1·4000 = 4000 mm

KL/r = 4000/7.5 = 533.3

Critical stress

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

Answer

σ_cr = 6.94 MPa

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