A girder that carries a uniformly distributed dead load of 1.7 k/ft plus its self-weight and three concentrated live loads of 15
kips at the quarter points of the 36 ft span is to be sized. Using A992 steel, determine the lightest Wshape to carry the load with lateral supports provided at the supports and load points. Use the correct Cb, check shear, and limit deflection to 1/360 of span.
1 answer:
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Answer:
- branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
- branch 2: i = 21.466∠63.435°; pf = 0.447 leading
- total: i = 31.693∠10.392° leading; pf = 0.984 leading
Explanation:
To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.
The inductive reactance is ...
The capacitive reactance is ...
<u>Branch 1</u>
The impedance of branch 1 is ...
Z1 = 8 +j4.99984 Ω
so the current is ...
I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°
The power factor is cos(-32.005°) ≈ 0.848 (lagging)
<u>Branch 2</u>
The impedance of branch 2 is ...
Z2 = 5 -j10 Ω
so the current is ...
I2 = 240/(5 +j10) ≈ 21.466∠63.435°
The power factor is cos(63.436°) ≈ 0.447 (leading)
<u>Total current</u>
The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.
It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)
It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°
The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)
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The phasor diagram of the currents is attached.
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<em>Additional comment</em>
Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.
Answer:
The work transfer per unit mass is approximately 149.89 kJ
The heat transfer for an adiabatic process = 0
Explanation:
The given information are;
P₁ = 1 atm
T₁ = 70°F = 294.2611 F
P₂ = 5 atm
γ = 1.5
Therefore, we have for adiabatic system under compression
Therefore, we have;
The p·dV work is given as follows;
Therefore, we have;
Taking air as a diatomic gas, we have;
The molar mass of air = 28.97 g/mol
Therefore, we have
The work done per unit mass of gas is therefore;
The work transfer per unit mass ≈ 149.89 kJ
The heat transfer for an adiabatic process = 0.
Answer:
Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib
Answer : moment of inertia = 186.7 Ib - in
Explanation:
Given data
weight of the mailbox = 3.2 Ib
weight of the uniform cross member = 10.3 Ib
The origin is of mailbox and cross member is 0
The perpendicular distance from Y axis of centroid of the mailbox
= 4 + (25/2) = 16.5"
The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"
therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)
= 52.8 + 133.9 = 186.7 Ib-in
