Answer:
Option A
Question (in proper order)
What is the correct expression of the internal moment in segment AB? A 10-ft-long cantilever beam is fixed at end A (x = 0). A uniform distributed load of 2 kip/ft is applied between A and B (x = 6 ft). A downward 10-kip force is applied at B, and an upward 8-kip force is applied at end C (x = 10 ft). In addition, a clockwise couple moment of 40 kip-ft is exerted at the free end C. What is the correct expression of the internal moment in segment AB?
(First Diagram attached below)
[tex]a. -x^{2} + 14x -56 kip.fit \\b. -x^2 + 14x kip.fit\\c. -x^2\\d. -x^{2} + 14x -24\\e. -x^{2} + 24\\f. -x^{2} - 56[/tex]
Explanation:
(From the second diagram attached below)
Note that
Moment of a force is the product of the force and the perpendicular distance
sum of horizontal forces, ∑Fx = 0 (that is sum of forces acting towards left = sum of forces acting towards right)
that is zero in this case
Sum of vertical forces, ∑Fy = 0 ( that is sum of forces acting upward = sum of forces acting downward)
Total upward force = Total downward force
Ay + 8 = 10 + 2 x 6
where Ay is the force acting at the fixed end
Ay + 8 = 10 + 12
Ay = 22 - 8
Ay = 14 Kip ( force is measured in kip while distance is measured in ft)
Sum of clockwise moment = Sum of anticlockwise moment
let us assume a counterclockwise moment about point A = Ma
8 x 10 + Ma = 40(counterclockwise moment) + (2 x 6) x 3(total uniform force assumed to act at the middle of the 6ft length) + 10 x 6
80 + Ma = 40 + 36 + 60
Ma = 136 - 80
Ma = 56 kip-fit
(From the third diagram attached below)
Consider a section of beam AB from A to X and of distance x fit
let the counterclockwise moment about Xx section be Mx
if we take moment about the Xx section
then
summation of moment about the Xx section ∑Mxx = 0
hence,
Mx + 56 + {(2 * x)[total of the uniform force] * x/2[acting at the middle point of the x ft distance]} = Ay * x
Mx + 56 + 2x * x/2 = 14 * x
Mx + 56 + x^2 = 14x
Mx = [tex]-x^2 + 14x - 56[/tex] kip-ft
