Structural Problem for AM

[geo pe]

Hello,

I cannot understand the logic behind multiplying DL with (13/12), see the problem attached. Any help will be much appreciated. Thanks!

Prob_1.pdf

 

[PEgeo]

Hello,
I cannot understand the logic behind multiplying DL with (13/12), see the problem attached. Any help will be much appreciated. Thanks!

It is the load on the horizontal plane, as follows

Total Dead Load = (WDL)L1/L2

where,

L1 = Slope length along the slope = 13

L2 = Horizontal length = 12

I found this in Design of wood structures book by Donald Breyer, the structural problems drive geos nuts at times!

 

[geo pe]

Thanks! makes sense now.

 

[darius]

Thanks! makes sense now.

I don't agree with the answer

RA = 30x16/2 + (20x12/13)x(16x13/12)/2=400

 

[darius]

Thanks! makes sense now.

I don't agree with the answer

RA = 30x16/2 + (20x12/13)x(16x13/12)/2=400

Could be right only if they would specify that 20#/ft is the perpendicular to beam component of the dead load of the slab, otherwise i would assume 20#/ft is the real weight of the slab.

 

[Jacob_PE]

For those of us not taking the structures afternoon exam, we don't have to worry about solving/analyzing indeterminite structures on the exam and so we can completely skip chapter 46 & 47 of the CERM (Structural Analysis I & II). True or False?

 

[Jacob_PE]

For those of us not taking the structures afternoon exam, we don't have to worry about solving/analyzing indeterminite structures on the exam and so we can completely skip chapter 46 & 47 of the CERM (Structural Analysis I & II). True or False?

Found an answer: http://engineerboards.com/index.php?showtopic=13512

 

[Jacob_PE]

Thanks! makes sense now.

I don't agree with the answer

RA = 30x16/2 + (20x12/13)x(16x13/12)/2=400

Could be right only if they would specify that 20#/ft is the perpendicular to beam component of the dead load of the slab, otherwise i would assume 20#/ft is the real weight of the slab.

I looked through my old college concrete and structures binders and couldn't find any examples that presented a dead load in this way. It seems to me they should have said flat out that the dead load is the weight of the slab or had the arrows within the slab. Normally we got a uniformally distributed load at an angle that, when applied over the length is the Resultant, not a subcomponent of the resultant. With this problem, there is no x component and y is the Resultant. Plus the problem starts out 'The problem has loading as figure.' Does that automatically tell you that the dead load applied is its weight?