Hinge's effect on uniform load on beam?

[StructureMan44]

While studying for the SE I came across problem 109 in the first attached document. The solution is added in the second post. My question is, how did they determine the 0.2 factor for the load due to the uniform load on the right side of support B? Does the hinge affect it; I thought hinges only affect moment transfer.

 

[StructureMan44]

Here is the solution to the problem.

 

[JAE]

The answer (without looking) is 46 kips correct? OK now I'll look.

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[JAE]

I'm not recalling how to build influence lines (been too long since I did bridges).

All I did was solve this non-redundant system and created the shear diagram after solving for the reactions.
The portion left of the hinge is a simple span which results in a point load on the end of the cantilever.

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[BAretired]

I don't particularly like using influence lines for this problem because the answer seems easier to obtain by direct analysis.

The triangle on the left span is the influence line for shear just to the right of Point B resulting from a unit load placed anywhere between points A and B. If the unit load is placed at A or B, it contributes nothing to moment in span A-B or shear at B.

If the unit load, say 1k is placed at the hinge, it produces a moment of 1*12 = 12'k at point B. This in turn produces equal and opposite reactions of 12/60 = 0.2 kips at B and C, upward at B and downward at C. The influence line is linear between the hinge and either A or B.

To find the effect of a uniform load between points A and B, you need to take the area under the influence line and multiply it by w, the uniform load.

BA

 

[IDS]

Like JAE I simply calculated the shear force at B for the UDL, and added the point load.

But if you want to use the influence line, calculate the shear force at B due to a unit point force at the hinge.
The span A-Hinge is simply supported, so a point load at the hinge will not transfer any load to A.

Take moments about C: 1 x 72
Reaction at B = 72/60 = 1.2
Shear force to right of B = 1.2 - 1 = 0.2

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[StructureMan44]

JAE: How were you able to traditionally solve this system? Did you break it into two beams at the hinge with a 24k reaction (up for the left beam, down for the right beam) or did you do something else?

 

[TLHS]

Yeah, that's the simple way to do it. By inspection the point load will be infinitesimally to the right of the section in question.

 

[JAE]

Yes - what TLHS said. Simple statics. With the hinge it is not a redundant system and therefore all you need is sum of forces Y and sum moments.

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[StructureMan44]

If we separate the beams the right beam has a 12ft cantilever on it's left side with a 24k downward force correct? Summing the moments about C we have:

EMc=o: 24kip(72ft)+10kip(60ft)-Rb(60ft)+72kip(36ft), Rb=82kip

Clearly this is the wrong approach?

 

[TLHS]

Why?

That gives you a shear of 46 kips to the right of the support.

Coming from the left, your shear starts at -24 kips from the point load and is driven to -36 kips by the distributed load by the time you reach the support. The support is 82 kips in the opposite direction, which means that immediately to the right of the support your beam shear switches direction and is 46kips for a very short distance before the moving point load shows up.

 

[Lion06]

This looks like too basic of a question to use an influence line. Whether by design (most likely) or not, the only item in question is where to place the concentrated load to maximize the internal shear on the right side of the support. It's easy to see by inspection where that would occur without constructing an influence line. If the intent of the question is to truly use an influence it would have been better to have a configuration that didn't require the uniform load be placed on the entire structure to achieve the maximum shear.

 

[jfmann]

As noted above by others........relatively simple just analyze segment to left of hinge first to obtain point load to be applied to end of cantilever.........then analyze remainder of beam as separate element (to right of hinge only). Of course if you were to consider moving load on the left-segment (which you do not for this problem, but just in general), you would have to calculate reaction at hinge with that load.

Showing solution using influence line is interesting..........but way too confusing to just arrive at solution!

John F Mann, PE

http://www.structural101.com

 

[StructureMan44]

If the influence line diagram in the solution is for shear, why is the highest point of the triangle on the left span not at support B? Unlike moment, shear can transfer through a hinge and doesn't "see" the hinge.

 

[Lion06]

It's because that's the location of load application that results in the highest shear force just to the right of Support B.

 

[BAretired]

SM44,
Lion06 is correct. A point load just left of support B produces zero shear just right of B as shown on the influence line in the left span. If that is not clear, think about Span B-C with a 12' cantilever and a concentrated to the left of Point B but very close to it. The shear to the left of B will be 10k but the shear to the right of B will be close to zero.

BA