Can someone explain how they came up with the correct number of attachment points for question #3, installing mechanical components? Thanks for your help.

7 replies

The answer given for this problem was wrong, here's how you do it:

module width is 39"

we allow 24" on each end

we need 1" between each and 3" on each end for mid/end clamps:

39" x 5= 195"

195" plus 3" at each end plus 4" between= 205"

205"- allowed 24" on each end" 205-24-24= 157"

Divide that by 48" allowed between each attacment:

157/48=

3.27

We must have no more than 4 spaces between modules, so 5 attachments are required per rail.

Total attachments= 20

if you click on the image you can see the full image. I don't know why it gets edged like this. Sorry.

module width is 39"

we allow 24" on each end

we need 1" between each and 3" on each end for mid/end clamps:

39" x 5= 195"

195" plus 3" at each end plus 4" between= 205"

205"- allowed 24" on each end" 205-24-24= 157"

Divide that by 48" allowed between each attacment:

157/48=

3.27

We must have no more than 4 spaces between modules, so 5 attachments are required per rail.

Total attachments= 20

if you click on the image you can see the full image. I don't know why it gets edged like this. Sorry.

[Last edited Mar 21, 2012 17:57:52]

Sarah, you are exactly right. I plugged in 62 for the module size and only figured for one row of 5. And, the rest is history. Just for practice, if you did have one row of 5 modules in landscape, would you agree that 7 attachments for each rail is right? But, Here is the corrected answer!!

39 x 5 + 4" (mid-clamps) + 6" (endclamps) = 205"

205 - 48 (to leave 24" on either end) = 157" (which is the maximum distance between the first and last attachments)

157 / 48 = 3.2 so I will have 3.2 48" spaces between the first and last attachment. If I go with 3, I will have more than 24" from the first/last attachments to the end, so I always have to round up, I can have less than 24", but never more, according to the manufacturer's instructions. So Lets actually map it out.

48 x 4 = 192

205 - 192 = 13 inches, so if I leave 6.5" from either end to the first attachment, I get the following attachment points plotted on a line:

End = 0, then 6.5, 54.5 (which is 6.5 + 48), 102.5 (54.5 + 48), 150.5 (102.5 + 48) and 198.5 (150.5 + 48). That is a total of (5) attachments per rail.

Each rail has (5) attachments, so (2) per row of 5 modules, or 10. For two rows, that is 10 x 2 = 20.

You will most likely see a question like this on the test, and they may or may not give guidance regarding mid- and end-clamp allowances. And, that could change your answer! In reality, often you can mount modules using the holes on the underide of the frame on, say Unistrut, but a question like this without knowing the manufacturer of the racking, is tough to interpret.

Be sure to follow the frame structure you are given (16", 20", or 24" on center), if they give you one, and I have never heard of a condition where more than 48" between attachments is acceptable. So, Good Luck! Remember, you can always write up the question, as long as you can justify your assumptions if you find more than one correct answer, depending on the clamp situation... Welcome to NABCEP!

39 x 5 + 4" (mid-clamps) + 6" (endclamps) = 205"

205 - 48 (to leave 24" on either end) = 157" (which is the maximum distance between the first and last attachments)

157 / 48 = 3.2 so I will have 3.2 48" spaces between the first and last attachment. If I go with 3, I will have more than 24" from the first/last attachments to the end, so I always have to round up, I can have less than 24", but never more, according to the manufacturer's instructions. So Lets actually map it out.

48 x 4 = 192

205 - 192 = 13 inches, so if I leave 6.5" from either end to the first attachment, I get the following attachment points plotted on a line:

End = 0, then 6.5, 54.5 (which is 6.5 + 48), 102.5 (54.5 + 48), 150.5 (102.5 + 48) and 198.5 (150.5 + 48). That is a total of (5) attachments per rail.

Each rail has (5) attachments, so (2) per row of 5 modules, or 10. For two rows, that is 10 x 2 = 20.

You will most likely see a question like this on the test, and they may or may not give guidance regarding mid- and end-clamp allowances. And, that could change your answer! In reality, often you can mount modules using the holes on the underide of the frame on, say Unistrut, but a question like this without knowing the manufacturer of the racking, is tough to interpret.

Be sure to follow the frame structure you are given (16", 20", or 24" on center), if they give you one, and I have never heard of a condition where more than 48" between attachments is acceptable. So, Good Luck! Remember, you can always write up the question, as long as you can justify your assumptions if you find more than one correct answer, depending on the clamp situation... Welcome to NABCEP!

Oops, this is an explanation to Question #2, not #1 or #3!

Now, for Question #3:

If you have actually 20 attachments, the answer here needs to change as well:

62” x 39”/144” = 16.77ft2

16.77 ft2 x 40 lbs/ft2 = 670.8 lbs.

670.8 lbs. x 10 modules = 6,708 lbs.

6708/20 attachments = 335.4 lbs./attachment

335.4/332lbs = 1.01” + 1” for decking = 2 1/4”, as lag screws typically come in 1/4" increments. If you could not find 2 1/4", then you'd go up to 2 1/2". Or, you can use the table figure 10-23 on page 274 of Dunlop text to choose a different diameter screw, so long as you do not exceed the 1.5 times the diameter of the bolt/width of truss rule (see SolarPro Article for discussion on this. GREAT article, by the way!!!)

If you have actually 20 attachments, the answer here needs to change as well:

62” x 39”/144” = 16.77ft2

16.77 ft2 x 40 lbs/ft2 = 670.8 lbs.

670.8 lbs. x 10 modules = 6,708 lbs.

6708/20 attachments = 335.4 lbs./attachment

335.4/332lbs = 1.01” + 1” for decking = 2 1/4”, as lag screws typically come in 1/4" increments. If you could not find 2 1/4", then you'd go up to 2 1/2". Or, you can use the table figure 10-23 on page 274 of Dunlop text to choose a different diameter screw, so long as you do not exceed the 1.5 times the diameter of the bolt/width of truss rule (see SolarPro Article for discussion on this. GREAT article, by the way!!!)

a little differntly:

62" x 39"= 2418" square

2418/ 144= 16.79 ft square

16.79 x 10= 167.9 ft square total

x 40 lb wind load= 6716 lb total load

6716 / 20 attachments= 335.8

tricky part:

335.8/332=1.01"

plus 1" for roof

= 2.01 or a 2 1/4" lag screw length

62" x 39"= 2418" square

2418/ 144= 16.79 ft square

16.79 x 10= 167.9 ft square total

x 40 lb wind load= 6716 lb total load

6716 / 20 attachments= 335.8

tricky part:

335.8/332=1.01"

plus 1" for roof

= 2.01 or a 2 1/4" lag screw length

If not specified, should you assume 1" for roof surface and decking?

Yes, 1" is standard due to usual 3/4" decking, plus tar paper, sealant, etc. so yes, 1". Of course that would be more if there was foam insulation or anything else!