Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic
Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.
Answer: 2.93 ft/sec
Explanation: Calculate the volume/sec entering from the two inlets (Pipes 1 and 2), add them, and then calculate the flow in Pipe 3.
The table illustrates the approach. I calculated the volume of each pipe for a 1 foot section with the indicated diameters, divided by 2 for the radius of each using V = πr²h. Units of V are in^3/foot length. Now we can multiply that volume by the flow rate, in ft/sec, to obtain the flow rate in in^3/sec.
Add the two rates from Pipes 1 and 2 (62.14 in^3/sec) to arrive at the flow rate for Pipe 3 necessary to keep the water level constant. Calculate the volume of 1 foot of Pipe 3 (21.21 in^3/foot) and then divide this into the inflow sum of 62.14 in^3/sec to find the flow rate of Pipe 3 (in feet/sec) necessary to keep the water level constant.
That is 2.93 ft/sec.
Answer:
Example for strengthening mechanism in single-phase material: Strain hardening- D.
Answer:
a)Bulk deformation process
Explanation:
<u>Rolling</u>
Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.
<u>Shearing</u>
In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.
So from above we can say that option a is right.
