Problem 1. A. Using singularity functions method, find out the equations for q, v, and M. B. Using appropriate boundary conditio
ns, find out the constants. C. Draw shear force and bending moment diagrams.
1 answer:
Answer:
Hello the required diagram for the question is missing attached is the diagram AND A FREE BODY DIAGRAM
B)Rav = 3.375 KN Rbv = 4.625 KN
Explanation:
Attached is the detailed diagram of the bending moment diagrams and shear force
A) using singularity functions
y(x) = Displacement shape of Elastic curve
dy/dx = ∅ (x) ( slope )
d^2x/dy^2 = m(x) / E I ( moment function )
d^3x / dy^3 = I / EI V(x) ( shear function )
d^4x / dy^4 = I / EI q(x) ( loading function )
B) using the appropriate boundary conditions find out the constants
summation of F(x) = 0
Rah = 0
Rav - 8 + Rbv = 0
Rav + Rbv = 8 ----- equation 1
summation of M(a) = 0
= -8(4) + Rbv (8) - 5 = 0
therefore Rbv = 37 / 8 = 4.625 KN
insert the value of Rbv back into equation 1
Rav = 8 - 4.635 = 3.375 KN
C) ATTACHED ARE THE DIAGRAMS
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Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑ = mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN