**Determine the shear stress in each section - cells:**

In Figure, the mean dimensions of the two cells are 100 mm × 50 mm and 50 mm square.

t_{1} = 3 mm, t_{2} = 6 mm, t_{3} = 3 mm

**Figure **

Determine the shear stress in each section, and the angle of twist per metre length for a torque of 3000 Nm. G = 80,000 N/mm^{2}.

**Solution**

A_{1} = (100 × 50) = 5000 mm^{2}

A_{2} = 50 × 50 = 2500 mm^{2}

Z_{1} = (2 × 100) + 50 = 250 mm

Z _{2} = (2 × 50) + 50 = 150 mm

Z_{3} = 50 mm

τ_{1} t_{1} = τ_{2} t_{2} + τ_{3} t_{3}

⇒ τ_{1} × 3 = τ_{2} × 6 + τ_{3} × 3

3 τ_{1} = 6 τ_{2} + 3 τ_{3 }---------- (1)

T = 2 (τ_{1} t_{1} A_{1} + τ_{2} t_{2} A_{2} )

⇒ 3000 × 10^{3} = 2 [τ_{1} × 3 × 3000 + τ_{2} × 6 × 2500]

⇒ τ_{1} + τ_{2} = 50 -------- (2)

(τ_{1} Z_{1 }+ τ_{3} Z_{3} ) / A_{1} = (τ_{2} Z_{2} - τ_{3} Z_{3} )/A_{2}

(τ_{1} × 250 + τ_{3 }× 50) /5000= (τ_{2} × 150 - τ_{3} × 50)/ 2500

⇒ 5 τ_{1} + τ_{3} = 6 τ_{2 } - 2 τ_{3}

⇒ 5 τ_{1} = 6 τ_{2} - 3 τ_{3} ------------ (3)

(3) - (1) ⇒ 2 τ_{1} = - 6 τ_{3}

τ_{1} = - 3 τ_{3 }---------- (4)

From (1),

3 (- 3τ_{3 }) = 6 τ_{2} + 3 τ_{3}

τ_{2 } = - 2 τ_{3}

From (2),

- 3 τ_{3} - 2 τ_{3} = 50 --------- (5)

τ_{3 } = - 10 N/mm^{2}

τ_{1} = - 3 (- 10) = 30 N/mm^{2}

τ_{2} = - 2 (- 10) = 20 N/mm^{2}

∴ θ= (τ_{1} Z_{1} + τ_{3 }Z_{3} ) l /2G A_{1} , Here l = 1000 mm

= [30 × 250 + (- 10) × 50]/ (2 × 80 × 103 × 5000) × 1000

= 8.75 × 10^{- 3} radians