Making use of the property established in Prob. 7.119, solve the problem indicated by first solv

Making use of the property established in Prob. 7.119, solve...

Key Concepts

Problem Reduction

This concept involves simplifying a complex problem by transforming it into an equivalent or similar problem that is already understood or easier to solve. The strategy leverages known results or structures from the simpler problem to address the original, more complicated problem.

Utilization of Established Properties

In mathematical problem solving, a powerful approach is to use previously proved properties or theorems to assist in solving new problems. By applying an established property, one can often bypass more cumbersome computations and directly reach a solution, ensuring the reasoning is built upon a solid foundation.

Solving Analogous Problems

This strategy entails first solving a corresponding or related problem whose solution is more straightforward, and then using that solution as a stepping stone to solve the original problem. It emphasizes the importance of recognizing similarities between problems so that insights gained from one context can be transferred to another.

Transcript

00:01 This is problem 122 of chapter 7 of vector mechanics for engineers, statics, and dynamics.

00:06 So in this problem, there is a cable going from a to e.

00:09 There is a load applied downward at b of two kips, at c of two kips, and at d of two kips.

00:16 The horizontal distance from a to b is six feet, from b to c is nine feet, from c to d is six feet, and from d to e is nine feet.

00:24 The vertical distance from e to a is 7 .5 feet.

00:28 The vertical distance from d to b, i'm sorry, b to e is distance b, from c to e is distance c, and from d to e is distance d.

00:40 And the problem we are given that the distance of c is 5 feet, and we're tasked with finding the distance of b and the distance of d.

00:48 So we need to solve this problem similar to how we solved problem 119.

00:53 So we're going to draw a simple beam underneath the forces, and we are going to start with doing the a cell of moments around e, setting it equal to zero.

01:05 This equals negative a, y, because it creates a negative moment around e, times 30 because there's 30 meters in the horizontal direction between a and e.

01:16 I'm going to add the force at b times 24, because 24 meters in the horizontal direction between b and e.

01:23 Add 2 times 15, 15 meters in the horizontal direction between c and e, and 2 times 9, it gives 9 meters between d and e in the horizontal direction.

01:35 So we can go ahead and solve for ay, which equals 3 .2 kips.

01:41 We can do the sum of forces in the y direction set equal to 0.

01:47 This is ay plus ey and subtracting all three forces.

01:53 Set equal to zero, we can solve for ey, which equals 2 .8 kips.

02:03 We now are going to solve for the shear forces.

02:06 Let's start with the sheer force of a, which is 3 .2, because it only involves a .y.

02:17 To the sheer force of b, which is 3 .2 minus the load at b.

02:23 So subtracting 2 equals 1 .2 cups.

02:28 Shear force at c equals 3 .2 minus load at b and minus the load at c.

02:35 It goes negative 0 .8.

02:38 Shear force at d goes the same thing.

02:41 Subtracting another load at d, which goes 2 .8, and then the sheer force add e, which is the same thing as a to d, and then we're going to add 2 .8, which is the force ey.

02:59 So we'll bring it back down to zero.

03:03 So i'm going to go ahead and graph that.

03:07 So this is 3 .2, and then 1 .2.

03:16 I'm going to go negative for the negative 0 .8.

03:21 And lastly, goes to the negative 2 .8, and then back down to 0.

03:30 Now we're going to calculate the bending moments.

03:35 So the bending moment at a equals the sheer force at a times the distance between a and b, which is 6 feet.

03:47 Oh, i'm sorry.

03:50 The bending moment at a starts off to be 0.

03:53 The bending moment at b is the sheer force of a times the distance between a and b, which is 6 feet...