Maximum slope of deflection of beam

[fonseh]

Homework Statement

why the author said that by inspection , the maximum slope occur at D ? How do we know that ? He didnt show the working and explanation .

Homework Equations

The Attempt at a Solution

If I consider the maximum slope to occur at a point after 2m from A , then i will take EI(dv2/dx2 ) = 0 , then my ans is , -2((x2)^2) + 12x2 -44/3 = 0 , then x = 4.29m ...

 

[fonseh]

Or it can be at both x = 4.29 m and x=1.63m ?

 

[Mapes]

Homework Statement

why the author said that by inspection , the maximum slope occur at D ? How do we know that ? He didnt show the working and explanation .

Do you mean the maximum deflection? The position of D is initially unknown but is simply assigned to that point.

If you're asking why the maximum deflection has to occur to the left of the load, draw a few examples with the load in different places and you'll see that unless the load is centered, the maximum deflection has to occur on the side where the load is farthest from a support.

 

[fonseh]

If you're asking why the maximum deflection has to occur to the left of the load, draw a few examples with the load in different places and you'll see that unless the load is centered, the maximum deflection has to occur on the side where the load is farthest from a support.

why ? i still can't imagine it

 

[Mapes]

why ? i still can't imagine it

Hmm, maybe a physical example would be helpful. Try supporting a long ruler at its ends A and C (with simple supports that allow rotation) and press down at different locations of B between the supports. If AB>BC, you'll never see the largest deflection occur within BC.

You may prefer a more rigorous, mathematical reason. Here's my attempt, which is unfortunately somewhat hand-wavy: the deflection is the distance integral of the bending moment, which is the distance integral of the shear. (Have you done shear and bending moment diagrams?) Because of these distance integrals, deflection is strongly enhanced with increasing distance. However, the ends are constrained to not deflect at all. With this constraint in mind, the farthest one can get from the load and the constrained ends is on the side of the beam without the downward load.

In your example, this is segment AB, and the author labels the lowest point D before calculating its location.

 

[fonseh]

However, the ends are constrained to not deflect at all. With this constraint in mind, the farthest one can get from the load and the constrained ends is on the side of the beam without the downward load.

In the case above , why shouldn't the maximum deflection occur at the location where P is applied ?

 

[Mapes]

There's no reason for that to occur. After all, if you hold a 1 m strip at x = 0 and push down at x = 0.5 m, will that be the location of maximum deflection? Of course not—it'll be at the unsupported end, at x = 1 m. As I wrote above, deflection increases strongly with distance from the load, barring any constraints.

 

[fonseh]

There's no reason for that to occur. After all, if you hold a 1 m strip at x = 0 and push down at x = 0.5 m, will that be the location of maximum deflection? Of course not—it'll be at the unsupported end, at x = 1 m. As I wrote above, deflection increases strongly with distance from the load, barring any constraints.

do you mean for the case above , the deflection will occur near to A because the distance AP is further than the distance of CP ?

 

[fonseh]

At here, we can notice that there are 2 sets of slope equation that we can use . Which is equation 5 & 7 . In equation 5 , we will get 1.633 as in the working . ( the author use by 'inspection' the max deflection occur at region AB) ? Why not

However , we don't know where is the position where the max deflection is located , how can we use equation 5 to solve ?

why We don't have to consider equation , which involve region DC ? By
solving equatio7= 0 , i have x = 5.23 , 3 and 0.763 , which is correct ?