1. How To Analyze Beams Mr. Ham Intro to Engineering

2. Now that you have your floor joist layout, it’s time to analyze our beams. 20’ 8’ 16’ 12’ 10-0-0 Take time now to dimension the location of your beams.

3. Now let’s look at our beam guide. We have three spans being held by the beam; a 20’ span, 28’ span, and 16’ span. Assuming we keep it one long beam, let’s find what our options are for 3-1/2” Microlam LVL bearing placement. (Let’s pick the biggest span of 28’ --- it’s better to over-engineer than under-engineer). 20’ 8’ 16’ 12’ We have three options:11 7/8” beam with bearings at every 12’ 14” beam with bearings at every 14’ 16” beam with bearings at every 16’ From a cost standpoint, the smaller the beam --- the cheaper the beam. From your standpoint, the fewer the spans, the fewer the calculations for you! We have three options:11 7/8” beam with bearings at every 12’ 14” beam with bearings at every 14’ 16” beam with bearings at every 16’

4. We have three options:11 7/8” beam with bearings at every 12’ 14” beam with bearings at every 14’ 16” beam with bearings at every 16’ Bearings every 12’ means we have: 50’/12’ = 4.17 spans, but we have to round up to 5. This results in 4 columns spaced 10’ apart. 20’ 8’ 16’ 12’ 20’ 25’ 5’ Span #1Span #2Span #3Span #4Span #5 Column #1 Column #2 Column #3 Column #4

5. 20’ 8’ 16’ 12’ 20’ 25’ 5’ Since the 14” beam and 16” beam both require 3 bearings, let’s go with the 14” beam. In conclusion, we will have a 14” beam --- carrying 4 spans --- with 3 bearings spaced 12’ 6” apart. 12’6”

6. 20’ 8’ 16’ 12’ 20’ 25’ 5’ My beam now looks like this: 12’6” 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE

7. Take time now to add columns and specify beam depth for all your floors. Then, draw a detailed drawing of each beam with beam locations as I did in the previous slide.

8. Floors are loaded at 40 PSF live load and 10 PSF dead load, for a total load of 50 PSF (pounds per square foot). 20’ 8’ 16’ 12’ 20’ 25’ 5’ 50 lbs. 50 lbs. each

9. Right now, we can assume that the beam holds ½ of each span. The other ½ of each span is supported by bearing walls. 20’ 8’ 16’ 12’ 5’ supported by wall (10’+8’)/2 = 9’ supported by wall 3’ supp. 5’ supported by wall 5’ supported by beam (10’+8’)/2 = 9’ supported by beam 3’ supp.

10. Now it’s time to turn our PSF (pounds per square foot) to a PLF (pounds per lineal foot). Remember, the floors are loaded at 50 PSF 5’ 8’ 16’ 12’ 5’ supported by wall (10’+8’)/2 = 9’ supported by wall 3’ supp. 5’ supported by wall 5’ supported by beam (10’+8’)/2 = 9’ supported by beam 3’ supp. 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE 20’ 25’ 5’ Each square is 50 pounds. 5’ 500 PLF Can you calculate this uniform load? ?? PLF Can you calculate this uniform load? 700 PLF 400 PLF Each foot of the left side of the beam is supporting 10 squares above. Each square is worth 50 pounds. So that means each foot of the left side of the beam is carrying 10 x 50 = 500 PLF

11. Our beam looks like scenario #2 and scenario #4. You might have some beams in your house similar to #5. Can you figure out the difference between scenario #2 & #4? Here the joists have been cut! Here the joists are continuous! We are going to use this scenario since it most closely resembles our floors. This is a phenomenon that occurs in engineering. Since the joists are continuous and not split over the beam… The BEAMS SUPPORT 5/8 of the span, not 1/2 as we assumed before. Therefore, we need to go back and change our numbers… We will multiply our loads by 1.25 to fix this. If you need to use scenario #5 for your design, you will multiply your loads by 1.10.

12. 5’ 8’ 16’ 12’ 5’ supported by wall (10’+8’)/2 = 9’ supported by wall 3’ supp. 5’ supported by wall 5’ supported by beam (10’+8’)/2 = 9’ supported by beam 3’ supp. 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE 20’ 25’ 5’ Each square is 50 pounds. 5’ 1.25 x 500 PLF 1.25 x 700 PLF 1.25 x 400PLF 625 PLF 875 PLF 500 PLF

13. 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE 1.25 x 500 PLF 1.25 x 700 PLF 1.25 x 400PLF 625 PLF 875 PLF 500 PLF Now it’s time to calculate our reactions at each column bearing. To do this, we must first split the uniform load at each bearing point. 625 PLF 20’ 25’ 5’ 875 PLF 500 PLF

14. L 6 = 500 lb/ft x 5 ft = 2500 lb L 5 = 875 lb/ft x 7.5 ft = 6562.5 lb 875 PLF L 4 = 875 lb/ft x 12.5 ft = 10937.5 lb L 3 = 875 lb/ft x 5 ft = 4375 lb L 2 = 625 lb/ft x 7.5 ft = 4687.5 lb L 1 = 625 lb/ft x 12.5 ft = 7812.5 lb 875 PLF 500 PLF 625 PLF 875 PLF 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE Now let’s turn each uniform load into a point load. Each point load will be located at the center of each uniform load. We need to also convert our feet/inches to decimals. 20’ 25’ 5’ 12.5’ Try converting these 4 loads on your ownSeriously, try these on your own I’ll wait… go get a calculator, pencil, and paper. SERIOUSLY?!?! Do you HONESTLY think just by clicking I’ll put the answers up on here? Well, fine, it’s your grade and not mine. Just don’t ask for help when you don’t know how to do it!

15. 2500 lb 6562.5 lb 10937.5 lb 4375 lb 4687.5 lb7812.5 lb 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE What we calculated are point loads, not uniform loads. The point load will be located at the center of its respective uniform load. 6.25’ 25’ 5’ 12.5’ 7812.5 lbs 4687.5 lbs 4375 lbs 10937.5 lbs 6562.5 lbs 2500 lbs 20’ 6.25’ 2.5’6.25’ 3.75’ 6.25’ 3.75’ 6.25’ 2.5’

16. 14” LVL beam 12’6” FAFA FBFB FCFC FDFD FEFE Now for your favorite part, we’ll take the sum of the moments around all 5 bearing points (pretending each span is its own isolated beam)! 1)M A = 0... -(6.25)(7812.5) + (12.5)(F B ) = 0 M B = 0… (6.25)(7812.5) - (12.5)(F A ) = 0 F A = 3906 lb, F B = 3906 lb 2) M B = 0... -(3.75)(4687.5) –(10)(4375) + (12.5)(F C ) = 0 M C = 0… (2.5)(4375) + (8.75)(4687.5) - (12.5)(F B ) = 0 F B = 4156 lb, F C = 4906 lb 3) M C = 0... -(6.24)(10937.5) + (12.5)(F D ) = 0 M D = 0… (6.25)(10937.5) - (12.5)(F C ) = 0 F C = 5469 lb, F D = 5469 lb 4) M D = 0... -(3.75)(6562.5) –(10)(2500) + (12.5)(F E ) = 0 M E = 0… (2.5)(2500) + (8.75)(6562.5) - (12.5)(F D ) = 0 F D = 5094 lb, F E = 3969 lb Now add up all of the forces: F A = 3906 lb, F B = 3906 + 4156 = 8062 lb, F C = 4906 + 5469 = 10375 lb, F D = 5469 + 5094 = 10563 lb, F E = 3969 lb 6.25’ 12.5’ 7812.5 lbs 4687.5 lbs 4375 lbs 10937.5 lbs 6562.5 lbs 2500 lbs 6.25’ 2.5’6.25’ 3.75’ 6.25’ 3.75’6.25’2.5’ Now let’s check our work… the forces up should equal the forces down. UP: 3906 lb + 8062 lb + 10375 lb + 10563 lb + 3969 lb = 36875 lb DOWN: 7812.5 lb + 4687.5 lb + 4375 lb + 10937.5 lb + 6562.5 lb + 2500 lb = 36875 lb SUCCESS!!!

17. Now refer to the previous slide and calculate the reaction at all of your bearing points. You MUST show ALL WORK on paper. Full credit will be awarded for analysis of all your house’s beams. Partial credit will be awarded for analysis of only 1 beam.