Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
Answer:
J
Explanation:
= Voltage of the battery = Potential difference across the plates of capacitor = 24 Volts
= Amount of charge being moved from one plate to another = 3 μC = 3 x 10⁻⁶ C
= amount of work done in moving the charge
Work done is given as
J
Complete Question:
Suppose , where A has the dimensions LT, B has dimensions L²T⁻¹, and C has dimensions LT². Then the exponents n and m have the values
Answer:
The value of n = ¹/₅
The value of m = ³/₅
Explanation:
Given dimensions;
A = LT
B = L²T⁻¹
C = LT²
The values of n and m are calculated as follows;