問題一覧
1
universal
2
countersunk
3
3/32
4
1/8
5
5/32
6
3/16
7
distance between the center of rivets
8
2D-4D
9
2.5D - 4D
10
height: .5D width: 1.5D
11
3/32= 0.09375 0.09375 x .5 = 0.046875 final answer .047”
12
3/32 = 0.046875 0.046875 x 3 = 0.140625 = 0.141 (rounded)
13
.125 x 32 = 4 Final answer 4/32
14
4/32 = 1/8” diameter = 1/8”
15
6/16 = 3/8” 3/8” final answer for length
16
6/32 = 3/16” 3/16” = final diameter
17
5/16 = 5/16” 5/16” is the final length
18
step 1. add materials given .032” + .032” + .040” = .104 .104 + ((1.5D))
19
the drill size + clecos in order goes SILVER 3 / 32” (in 32nds for first and third cleco) COPPER 1 / 8” ( only one in 8ths) BLACK 5 / 32” GOLD 3 / 16” (only one in 16ths)
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Zachary Halycz · 54問 · 2年前Composites test 2
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Structures test 2, memorization
11問 • 2年前問題一覧
1
universal
2
countersunk
3
3/32
4
1/8
5
5/32
6
3/16
7
distance between the center of rivets
8
2D-4D
9
2.5D - 4D
10
height: .5D width: 1.5D
11
3/32= 0.09375 0.09375 x .5 = 0.046875 final answer .047”
12
3/32 = 0.046875 0.046875 x 3 = 0.140625 = 0.141 (rounded)
13
.125 x 32 = 4 Final answer 4/32
14
4/32 = 1/8” diameter = 1/8”
15
6/16 = 3/8” 3/8” final answer for length
16
6/32 = 3/16” 3/16” = final diameter
17
5/16 = 5/16” 5/16” is the final length
18
step 1. add materials given .032” + .032” + .040” = .104 .104 + ((1.5D))
19
the drill size + clecos in order goes SILVER 3 / 32” (in 32nds for first and third cleco) COPPER 1 / 8” ( only one in 8ths) BLACK 5 / 32” GOLD 3 / 16” (only one in 16ths)