Given:
Reynolds number,
Re = 1000As we know the formula,
→ Friction factor, [tex]F = \frac{64.00}{Re}[/tex]
By substituting the values, we get
[tex]= \frac{64.00}{1000}[/tex]
[tex]= 0.064[/tex]
Thus the response above is correct.
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Two stepped bar is supported at both ends.At the join point of two segments,the force F is applied(downwards).Calculate reactive forces R1 and R3 at the supports.What is value of absolute maximal stress?
Choose one answer nearest your result.
given= d1=10mm d2=20mm L1=20mm L2=10mm E=200GPa F=20kN
Answer:
F=200kN
Explanation:
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P
Answer:
hello a diagram attached to your question is missing attached below is the missing diagram
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m
Answer :
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa
Explanation:
Given data:
Type of steel = A-36
cross-sectional area = 500 mm^2
Calculate the average normal stress in each bar
we have to make some assumptions
assume forces in AB, CD, EF to be p1,p2,p3 respectively
∑ Fy = 0 ; p1 + p2 + p3 = 70kN ---------- ( 1 )
∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0
where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )
Take ; Tan∅
Tan∅ = MN / 2d = OP/d
i.e. s1 - 2s2 - s3 = 0
[tex]\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0[/tex]
L , E and A are the same hence
P1 - 2p2 + p3 = 0 ----- ( 3 )
Next resolve the following equations
p1 = 40.03 kN, p2 = 23.33 kN, p3 = 5.33 kN
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa
A universal shift register can shift in both the left-to-right and right-to-left directions, and it has parallel-load capability. Draw a circuit for such a shift register.
Answer:
Explanation:
A unidirectional shift register allows for the capability of shifting in one direction as the name unidirectional implies.
A bidirectional shift register has the capabilities of shifting in both the left to right and right to left directions.
For a Universal shift register, there are possibilities of bidirectional shifts and parallel-load capabilities that have the following properties.
There is the existence of a clear control input whose main FUNCTION is to set all the register to 0.'
A shift control to both the right and left direction to enable both the shift right operation and shift left operation.
Finally, a parallel-load control whose function is to activate a parallel transfer from input to output.
The diagram for the circuit for such a shift register can be seen in the diagram attached below.