Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:A composite consists of short SiC fibers embedded in an Al matrix, tested under isostrain. The fib

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:A composite consists of short SiC fibers embedded in an Al matrix, tested under isostrain. The fibers have a tensile strength of 
2
0
0
0
 ﻿MPa and a diameter of 
0
.
0
1
 ﻿mm
.
 ﻿The Al matrix, at the relevant strain, carries 
1
5
0
 ﻿MPa when composite fails. The interfacial shear strength is 
5
0
 ﻿MPa, the fiber length is 
2
.
0
 ﻿mm
,
 ﻿and the fiber volume fraction is 
4
0
%
.
 
1
.
 ﻿The critical fiber length? 
2
.
 ﻿Estimate the composite tensile strength. 
3
.
 ﻿Ask above question if fiber length is 
0
.
1
 ﻿mm
.

VIPSolutions · Accepted Answer

# Composite Material Problem (Si Fibers in Matrix) Given: - **Fiber tensile strength** ($\sigma_f$): $200~\text{MPa}$ - **Fiber diameter** ($d$): $.01~\text{mm}$ - **Matrix stress at composite failure** ($\sigma_m$): $150~\text{MPa}$ - **Interfacial shear strength** ($\tau_i$): $50~\text{MPa}$ - **Fiber length** ($l$): $2.~\text{mm}$ (also consider $.1~\text{mm}$) - **Fiber volume fraction** ($V_f$): $40\%$ or $.4$ --- ## 1. **Critical Fiber Length ($l_c$)** The **critical fiber length** is the minimum length required so the fiber can reach its maximum strength before pulling out. $$ l_c = \frac{\sigma_f \cdot d}{2 \tau_i} $$ **Plug in the values:** - $\sigma_f = 200~\text{MPa}$ - $d = .01~\text{mm}$ - $\tau_i = 50~\text{MPa}$ $$ l_c = \frac{200 \times .01}{2 \times 50} $$ $$ l_c = \frac{20}{100} = .2~\text{mm} $$ **Final answer:** > **Critical fiber length, $l_c = .2~\text{mm}$** --- ## 2. **Estimate Composite Tensile Strength ($\sigma_{c}$)** **Case:** Fiber length ($l = 2.~\text{mm}$) **is much greater than critical length ($l_c = .2~\text{mm}$)**. - Under **isostrain** (load applied parallel to fibers; both phases deform equally). **Rule of Mixtures (assuming fibers reach full strength):** $$ \sigma_{c} = V_f \sigma_f + V_m \sigma_m $$ Where: - $V_f = .4$ - $V_m = 1 - V_f = .6$ - $\sigma_f = 200~\text{MPa}$ - $\sigma_m = 150~\text{MPa}$ **Calculation:** $$ \sigma_{c} = .4 \times 200 + .6 \times 150 $$ $$ = 800 + 90 = 890~\text{MPa} $$ **Final answer:** > **Composite tensile strength, $\sigma_{c} = 890~\text{MPa}$** --- ## 3. **If Fiber Length is $.1~\text{mm}$ (i.e., $l < l_c$):** Now, the fiber is **shorter than critical length**. It cannot reach its full tensile strength before pulling out. ### **Maximum strength the fiber can contribute:** The maximum stress in short fiber ($\sigma_f'$) is: $$ \sigma_f' = \frac{2 \tau_i l}{d} $$ Plug in: - $\tau_i = 50~\text{MPa}$ - $l = .1~\text{mm}$ - $d = .01~\text{mm}$ $$ \sigma_f' = \frac{2 \times 50 \times .1}{.01} = \frac{10}{.01} = 100~\text{MPa} $$ ### **Composite strength:** $$ \sigma_{c} = V_f \sigma_f' + V_m \sigma_m $$ $$ = .4 \times 100 + .6 \times 150 = 400 + 90 = 490~\text{MPa} $$ **Final answer:** > **Composite tensile strength with short fibers ($l = .1~\text{mm}$): $\sigma_{c} = 490~\text{MPa}$** --- ## **Summary Table** | Quantity | $l = 2.~\text{mm}$ | $l = .1~\text{mm}$ | |---------------------------|:---------------------:|:---------------------:| | Critical length, $l_c$ | $.2~\text{mm}$ | $.2~\text{mm}$ | | Composite strength | $890~\text{MPa}$ | $490~\text{MPa}$ | --- ## **Key Takeaways** - **Critical fiber length** is the threshold above which fibers can fully utilize their strength. - **Long fibers ($l > l_c$)**: Use rule of mixtures with full fiber strength. - **Short fibers ($l < l_c$)**: Use modified strength limited by interfacial shear. --- **If you have more questions on composite design, let me know!**

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