Maximum excess pressure inside a thin-walled steel tube of radius r and thickness `/_\r (lt lt r)`, so that the tube would not rupture would be (breaking stress of steel is `sigma_("max")`

Options

A. `sigma_("max")xxr/(/_\r)`
B. `sigma_("max")xx(/_\r)/r`
C. `sigma_("max")`
D. `sigma_("max")xx(/_\2r)/r`

Explanation

To solve the problem of finding the maximum excess pressure inside a thin-walled steel tube, we can follow these steps:

### Step 1: Understand the Geometry and Forces
We have a thin-walled steel tube with radius \( r \) and thickness \( \delta r \) (where \( \delta r \ll r \)). The tube experiences internal pressure, leading to tension in the walls of the tube.

### Step 2: Draw a Free Body Diagram
Draw a cross-section of the tube and identify the forces acting on it. The tension \( T \) in the tube walls acts tangentially. The angle subtended at the center by the tension is \( 2\theta \).

### Step 3: Apply Force Equilibrium
Using the equilibrium of forces, we can express the relationship between the tension and the pressure difference. The vertical components of the tension must balance the pressure acting on the area of the tube.

The force due to the internal pressure \( \Delta P \) acting on the area of the tube can be expressed as:
\[ \text{Force due to pressure} = \Delta P \times (2r \cdot l) \]
where \( l \) is the length of the tube.

### Step 4: Relate Tension and Pressure
From the geometry, the vertical component of the tension can be expressed as:
\[ 2T \sin(\theta) \approx 2T\theta \quad (\text{for small } \theta) \]
Setting the two forces equal gives:
\[ 2T\theta = \Delta P \times (2r \cdot l) \]
Cancelling the common factors (2 and \( \theta \)):
\[ T = \Delta P \cdot r \cdot l \]

### Step 5: Relate Tension to Stress
The breaking stress \( \sigma_{\text{max}} \) of the material is defined as:
\[ \sigma_{\text{max}} = \frac{T}{\delta r \cdot l} \]
Substituting for \( T \) from the previous step:
\[ \sigma_{\text{max}} = \frac{\Delta P \cdot r \cdot l}{\delta r \cdot l} \]
Cancelling \( l \) from both sides:
\[ \sigma_{\text{max}} = \frac{\Delta P \cdot r}{\delta r} \]

### Step 6: Solve for Maximum Excess Pressure
Rearranging the equation to solve for \( \Delta P \):
\[ \Delta P = \sigma_{\text{max}} \cdot \frac{\delta r}{r} \]

### Final Result
Thus, the maximum excess pressure inside the thin-walled steel tube is given by:
\[ \Delta P = \sigma_{\text{max}} \cdot \frac{\delta r}{r} \]

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