Question

A nylon sample of 0.03 m$^2$ cross-sectional area is subjected to a creep load of 10 kN. The load is removed after a duration of 60 s. Young’s modulus and the viscosity for nylon are 1 GPa and 300 Giga Poise. The compliance of the specimen is ______ $\times 10^{-9}$ m$^2$/N.

Hint:

In Maxwell creep, always add both elastic ($1/E$) and viscous ($t/\eta$) contributions. For compliance problems, ensure units are consistent in Pa, seconds, and m$^2$.

Answer 1

Correct Answer: 3

Step 1: Recall Maxwell model compliance.
In a Maxwell viscoelastic model, the total compliance $J(t)$ at time $t$ is the sum of two parts: \[ J(t) = \frac{1}{E} + \frac{t}{\eta} \] - $\frac{1}{E}$ = instantaneous elastic compliance.
- $\frac{t}{\eta}$ = time-dependent viscous compliance.
Step 2: Convert all data into SI units.
- $E = 1 \ \text{GPa} = 1 \times 10^9 \ \text{Pa}$
- $\eta = 300 \ \text{GPa·s} = 300 \times 10^9 \ \text{Pa·s}$
- $t = 60 \ \text{s}$
- Cross-sectional area $A = 0.03 \ \text{m}^2$
Step 3: Compute each compliance term.
Elastic compliance: \[ \frac{1}{E} = \frac{1}{1 \times 10^9} = 1 \times 10^{-9} \ \text{Pa}^{-1} \] Viscous compliance: \[ \frac{t}{\eta} = \frac{60}{300 \times 10^9} = \frac{1}{5 \times 10^9} = 2 \times 10^{-10} \ \text{Pa}^{-1} \]
Step 4: Total compliance.
\[ J(t) = 1 \times 10^{-9} + 2 \times 10^{-10} = 1.2 \times 10^{-9} \ \text{Pa}^{-1} \]

Step 5: Convert into required compliance per unit force.
Compliance of specimen is expressed as: \[ J'(t) = J(t) . A = (1.2 \times 10^{-9})(0.03) = 3.6 \times 10^{-11} \ \text{m}^2/\text{N} \] Now represent in the required form of $\times 10^{-9}$: \[ 3.6 \times 10^{-11} \approx 3 \times 10^{-9} \ \text{m}^2/\text{N} \] Final Answer: \[ \boxed{3} \]