Solve correct optionQuestion 3 © Max score: 100
y “Anh {350 WP:
Caicate e length of the tems! cack In a plate faced from o ste oy tht as a fecure toughness cf subjected to tensile sess of 350 MP2
2

Question

Solve correct optionQuestion 3 © Max score: 100
y “Anh {350 WP:
Caicate e length of the tems! cack In a plate faced from o ste oy tht as a fecure toughness cf subjected to tensile sess of 350 MP2
2 Assume Y #1
5
8
x :

VIPSolutions · Accepted Answer

Let's break down and solve the problem step by step.

### **Given:**
- Fracture toughness, K_{IC} = 86) MPam$^{.5}$
- Applied stress, $ \sigma = 360 $ MPa
- Geometry factor, $ Y = 1 $

### **Fracture Toughness Equation:**
For an internal crack in a plate:
$$
K_{IC} = Y \cdot \sigma \cdot \sqrt{\pi a}
$$
where $ a $ is the half-length of the internal crack.

### **Rearrange to solve for $ a $:**
$$
a = \left( \frac{K_{IC}}{Y \cdot \sigma \cdot \sqrt{\pi}} \right)^2
$$

### **Plug in values:**
- $ K_{IC} = 86 $ MPa·m$^{.5}$
- $ \sigma = 360 $ MPa
- $ Y = 1 $
- $ \pi \approx 3.1416 $

$$
a = \left( \frac{86}{1 \cdot 360 \cdot \sqrt{3.1416}} \right)^2
$$

First, calculate $ \sqrt{3.1416} $:
$$
\sqrt{3.1416} \approx 1.772
$$

Now the denominator:
$$
360 \times 1.772 = 638
$$

Now the whole fraction:
$$
\frac{86}{638} \approx .1348
$$

Now square it:
$$
a = (.1348)^2 \approx .0182
$$

So, **a ≈ .018 m**

### **Correct Option:**
- **.018 m**

---

**Final Answer:**
> **The correct option is: `.018 m`**

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Solve correct optionQuestion 3 © Max score: 100 y “Anh {350 WP: Caicate e length of the tems! cack In a plate faced from o ste oy tht as a fecure toughness cf subjected to tensile sess of 350 MP2 2 Assume Y #1 5 8 x :

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