Answer to Question #229142 in Molecular Physics | Thermodynamics for Unknown346307

Question

A thick cylinder of 100 mm external diameter and 50 mm internal diameter is wound with

steel wire of 1 mm diameter, initially stressed to 20 MN/m’ until the outside diameter is

120 mm. Determine the maximum hoop stress set up in the cylinder if an internal pressure of

30 MN/m’ is now applied.

Answer 1

Given,

External diameter $(R)=100 mm=100\times 10^{-3}m$

Internal diameter $(r)= 50 mm=50\times 10^{-3}m$

$r_1=0.025$

Diameter of steel wire $(d)=1mm=10^{-3}m$

Applied stress$= 20MN/m =20\times 10^6N/m$

Outside diameter $(d)=120mm =120\times 10^{-3}m$

$(r_2)=0.06m$

Internal pressure $(\sigma_r)=30MN/m = 30\times 10^6N/m$

$\Rightarrow -30=A-\frac{B}{0.025}=A-1600B...(i)$

$\Rightarrow 0=A-\frac{B}{0.06^2}=A-278B ...(ii)$

From equation (i) and (ii),

$\Rightarrow 30=1322B$

$\Rightarrow B=\frac{30}{1322}=0.0227$

$\Rightarrow A=278B$

$\Rightarrow A=278\times 0.0227 =6.32$

For calculating the stress, the wire and cylinder should be treated like a single thick cylinder with internal diameter 50mm and 120mm external diameter.

Hoop stress at 25mm is $(\sigma_{r_1})=A+1600B=42.7MN/m^2$

Hoop stress at 60mm is $(\sigma_{r_2})=15.4MN/m^2$

$\Rightarrow \sigma_r=-\frac{r^2-R^2}{2r^2}T\log(\frac{R_3^2-R_1^2}{r^2-R_1^2})=p$

$\Rightarrow \sigma_r=\frac{0.05^2-0.025^2}{2\times 0.05^2}T\log_e(\frac{0.06^2-0.025^2}{0.05^2-0.025^2})$

$\Rightarrow \sigma_r = \frac{18.75\times 20}{50}\log_e(\frac{29.75}{18.75})$

$\Rightarrow \sigma_r=-3.45MN/m^2$

So, for the wire winding, the stress in the tube

$\sigma_r=-3.45$ at $r=0.05$

and $\sigma_r =0$ to $r=0.025$

$\sigma _r=A-400B$

$\Rightarrow -3.45=A-400B ....(iii)$

and $0=A-1600B...(iv)$

Now, solving equation (iii) and equation (iv)

$-3.45=1200B$

$\Rightarrow B=2.88\times 10^{-3}$ and $A=-4.6$

Hoop stress at 25mm is $(\sigma_{r_1})=A+1600B=-9.2MN/m^2$

Hoop stress at 50mm is $=A+400B =-5.75 MN/m^2$

Hence, the internal pressure at

r=25mm, $\sigma_H=42.7-9.2=33.5MN/m^2$

$r=50mm,$ $\sigma_H=15.4-5.75 =9.65MN/m^2$

So, the maximum hoop stress $33.5MN/m^2$