The question is incomplete. The complete question is :
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:
J(t) = Jo (1 - exp (-t/to))
Where,
Jo= 3m^2/ GPA
to= 200s
Determine
a) the strain after 100's (before stress is reversed)
b) the residual strain when stress falls to zero.
Answer:
a)-60GPA
b) 0
Explanation:
Given t= 0,
σ = 20Mpa
Change in σ= 0.2Mpas^-1
For creep compliance material,
J(t) = Jo (1 - exp (-t/to))
J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa
a) t= 100s
E(t)= ΔσJ (t - Jo)
= 0.2 × 3 ( 100 - 200 )
= 0.6 (-100)
= - 60 GPA
Residual strain, σ= 0
E(t)= Jσ (Jo) ∫t (t - Jo) dt
3 × 0 × 200 ∫t (t - Jo) dt
E(t) = 0
Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Metals are malleable and ductile.
Metals are good conductors of heat and electricity.
Metals are lustrous (shiny) and can be polished.
Metals are solids at room temperature (except mercury, which is liquid).
Metals are tough and strong.
hope this helps!
