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

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

P = 110 kNd = 10 mmL = 2 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

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

P = 55 kNsegment 1: Ø25 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, brass (E = 100 GPa)

δ = mm

3.Torsion of shaftsMedium

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 = 2 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1500 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 9 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 2 kN/m from 0 to 10 mx = 9 m

M_x = kN·m

6.Shear force & bending momentHard

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

L = 4 mP = 25 kN at 2 mw = 2 kN/m from 0 to 4 mx = 2 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 150 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 150 mmd = 40 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 168 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmcut-out: 50 × 40 mmhole centre: y = 168 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 80 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 80 MPaσ_y = −60 MPaτ_xy = 10 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (σ_z = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 160 MPaσ_y = 0 MPaτ_xy = 10 MPaσ_Y = 250 MPa

n =

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 0.5 m from A. Find the deflection δ_P under the load.

P = 21 kNa = 0.5 mL = 4 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 3.5 mE = 11 GPa (timber)b = 125 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)b × h = 25 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 60 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 60 kNn = 3L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 7003 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·10²/4 = 78.54 mm²

σ = P/A = 110·1000/78.54 = 1401 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 1401/200000 = 0.007003 = 7003 µε

Answer

ε = 7003 µε

2.Axial stress & strain

Answer: δ = 6.544 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 = 112 MPa

σ₂ = 486.3 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 112·1500/100000 = 1.681 mm

δ₂ = σ₂L₂/E₂ = 486.3·1000/100000 = 4.863 mm

δ = δ₁ + δ₂ = 6.544 mm

Answer

δ = 6.544 mm

3.Torsion of shafts

Answer: φ = 3.695 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·2000/(39000·79522) = 0.06449 rad

φ = 0.06449·180/π = 3.695°

Answer

φ = 3.695°

4.Torsion of shafts

Answer: d = 37.81 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 100000/157.1 = 636.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·636620/(π·60)) = 37.81 mm

Answer

d = 37.81 mm

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

5.Shear force & bending moment

Answer: M_x = 9 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 = 2·10 = 20 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 2·10·5

R_B = 100/10 = 10 kN

ΣF_y = 0: R_A = 20 − 10 = 10 kN

Cut at x = 9 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 = 10·9 − 2·9·4.5 = 9 kN·m

Answer

M_x = 9 kN·m

6.Shear force & bending moment

Answer: M_x = 29 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·4 = 8 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 25·2 + 2·4·2

R_B = 66/4 = 16.5 kN

ΣF_y = 0: R_A = 33 − 16.5 = 16.5 kN

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

M_x = 16.5·2 − 2·2·1 = 29 kN·m

Answer

M_x = 29 kN·m

7.Centroid & second moment of area

Answer: I_x = 55.46 × 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 = (30000·75 − 1257·52.5)/(30000 − 1257) = 75.98 mm

Full plate about ȳ

200·150³/12 + 30000·−0.9837² = 56.28 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·−23.48² = 0.8187 × 10⁶ mm⁴

Subtract

I_x = 55.46 × 10⁶ mm⁴

Answer

I_x = 55.46 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 225.3 × 10⁶ mm⁴

Centroid

ȳ = (48000·120 − 2000·168)/(48000 − 2000) = 117.9 mm

Full plate about ȳ

200·240³/12 + 48000·2.087² = 230.6 × 10⁶ mm⁴

Hole about ȳ

50·40³/12 + 2000·50.09² = 5.284 × 10⁶ mm⁴

Subtract

I_x = 225.3 × 10⁶ mm⁴

Answer

I_x = 225.3 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −40.93 °

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 10/70.71 = 0.1414, sin 2θ_s = −70/70.71 = −0.9899

2θ_s = atan2(−70, 10) = −81.87°

Angle on the element

θ_s = −40.93°

check: θ_s − θ_p = −40.93 − 4.065 = −45°

Answer

θ_s = −40.93°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 1.55

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 σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 80 MPa, R = 80.62 MPa

σ₁ = 160.6 MPa, σ₂ = −0.6226 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 161.2 MPa

n = 250/161.2 = 1.55

Answer

n = 1.55

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

11.Beam deflection

Answer: δ_P = 1.396 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 = 21000 N, a = 500 mm, L = 4000 mm

EI = 6.27 × 10¹¹ N·mm²

Deflection under the load

δ_P = Pa³/(3EI) = 21000·500³/(3·6.27 × 10¹¹) = 1.396 mm

Answer

δ_P = 1.396 mm

12.Beam deflection

Answer: h_req = 144.9 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.

w = 2.5 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 3500/250 = 14 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·3500⁴/(384·11000·14) = 3.172 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·3.172 × 10⁷/125)^1/3 = 144.9 mm

Answer

h_req = 144.9 mm

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

13.Column buckling

Answer: P_cr = 110.2 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·78125/700² = 110152 N = 110.2 kN

Answer

P_cr = 110.2 kN

14.Column buckling

Answer: d = 67.88 mm

Required critical load

P_cr = n·P = 3·60000 = 180000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·180000·2000²/(π³·70000))^(1/4) = 67.88 mm

Answer

d = 67.88 mm

Check: the slenderness KL/r is 117.9, so the column is slender and Euler applies.

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