9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 * 10⁷ Pa? Assume that each rivet is to carry one quarter of the load.

If $\mu$  is Poisson’s ratio,

Y = 3K (1 - 2 $\mu$ )      ……… (1)

and Y = 2 $n\left(1+\mu \right)$  ……… (2)

With the help of equations (1) and (2), we can write

  $\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
0.5 × (5×10²) × 1.6×10? ³ = 20×10? ³ v²
0.4 = 20×10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 2

As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$