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2 years ago

6

In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.

8) is 1.2 and there may be a 2 mm crack present in a bone specimen, what would be the maximum tensile stress that can be applied before fracture occurs?

1 answer:

Diano4ka-milaya [45]2 years ago

6 0

Where is Eq.(28) ?? You should show it to find the result

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The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

8 0

2 years ago

What can be used to relieve stress in a weld.

satela [25.4K]

Yes , of course you can

3 0

2 years ago

Read 2 more answers

Joinn my zo om lets play some blookets<br> 98867 708157<br> 9dPQPW

pav-90 [236]

Answer:k

Explanation:

6 0

2 years ago

A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of

zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

So Total net Force on each bolt Fb = P_Eb + Pi

Fb = 4.9 kip + 40 kip

Fb = 44.9 kip

Therefore the net force in every bolt is 44.9 kip

4 0

2 years ago

6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

zzz [600]

Answer:

Heat flux of CO₂ in cgs

= 170.86 x 10⁻⁹ mol / cm²s

SI units

170.86 x 10⁻⁸ kmol/m²s

Explanation:

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2 years ago