The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).
<h3>What is the distribution of pressure at some stage in bending?</h3>
Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.
The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.
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The resistance and voltage drop will still increase but at a smaller rate than the intended axis such as the long axis.
Width * Length * Thickness * Density = Weight.
48″ * 96″ * . 1875″ * 0.284 lb/in3 = 245 lb.
Answer:
The answer is 960 kg
Explanation:
Solution
Given that:
Assume the initial dye concentration as A₀
We write the expression for the dye concentration for one hour as follows:
ln (C₁) = ln (A₀) -kt
Here
C₁ = is the concentration at 1 hour
t =time
Now
Substitute 480 g for C₁ and 1 hour for t
ln (480) = ln (A₀) -k(1) ------- (1)
6.173786 = ln (A₀) -k
Now
We write the expression for the dye concentration for three hours as follows:
ln (C₃) = ln (A₀) -k
Here
C₃ = is the concentration at 3 hour
t =time
Thus
Substitute 480 g for C₃ and 3 hour for t
ln (120) = ln (A₀) -k(3) ------- (2)
4.787492 = ln (A₀) -3k
Solve for the equation 1 and 2
k =0.693
Now
Calculate the amount of blue present initially using the expression:
Substitute 0.693 for k in equation (2)
4.787492 = ln (A₀) -3 (0.693)
ln (A₀) =6.866492
A₀ =e^6.866492
= 960 kg
Therefore, the amount of the blue dye present from the beginning is 960 kg
Answer:
Your question is lacking some information attached is the missing part and the solution
A) AB = AD = BD = 0, BC = LC
AC = 
B) AB = AD = BC = BD = 0
AC = 
Explanation:
A) Forces in all members due to the load L in position A
assuming that BD goes slack from an inspection of Joint B
AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C
B) steps to arrive to the answer is attached below
AB = AD = BC = BD = 0
AC = 