Answer:
True
Explanation:
For point in xz plane the stress tensor is given by![\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DDx_%7B%7D%20%26txz%5C%5Ctzx%26Dz%5C%5C%5Cend%7Barray%7D%5Cright%5D)
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane
Answer:
Explanation:
Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.
Answer:
Benefits that last a lifetime. The Army offers you money for education, comprehensive health care, generous vacation time, family services and support groups, special pay and cash allowances to cover the cost of living.
Explanation:
Outer di ameter ![d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}](https://tex.z-dn.net/?f=d_%7B0%7D%3D400%20%5Cmathrm%7Bmm%7D%5Btex%5D%20Thickness%20of%20the%20cylinder%20%5Btex%5Dt%3D10%20%5Cmathrm%7Bmm%7D)
![\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10](https://tex.z-dn.net/?f=%5Ctherefore%5Btex%5D%20Inner%20diam%20eter%20%5Btex%5Dd_%7Bi%7D%3Dd_%7B0%7D-2%20t%3D400-2%20%5Ctimes%2010)
Given loading on the cylinder 
Helix an gle of the weld form 
(i) Normal stress on the plane at angle is
(ii) Shear stress along an angle of is 
Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
