Total Number Gearing Ratio
a.)
To determine the total gearing ratio we know
𝑛𝑜𝑢𝑡
𝑚𝐺,𝑡𝑜𝑡 =
𝑛𝑖𝑛
= 𝑚𝐺23 × 𝑚𝐺45
were 𝑚𝐺23= 5, 𝑚𝐺45= 2, 𝑛𝑜𝑢𝑡= 1000 rpm, and 𝑛𝑖𝑛= 100 rpm
As a result we get, 𝑚𝐺,𝑡𝑜𝑡= 10
Finding N2, N3, N4, N5 with appropriate P23 and P45
b.)
Using pressure angle ϕ = 20° and given center distance ds= 6 in, and
mG23 = 5
We know,
𝑁
𝑁
= 𝑁𝐺,𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑁3 = 5
𝐺,23
1.) 𝑚
𝑃,𝑎𝑐𝑡𝑢𝑎𝑙
2
To avoid interference accordingly for spur gears (k=1, full depth teeth), we
start by find minimum teeth that pinion can have using equation 2
2.)𝑁
=
𝑃,𝑚𝑖𝑛
⎡
⎢𝑚
+
2
(1+2𝑚𝐺,23)𝑠𝑖𝑛 (ϕ) ⎢ 𝐺,23
⎣
⎤
2
2
𝑚 𝐺,23 + (1 + 2𝑚𝐺,23)𝑠𝑖𝑛 (ϕ)⎥
⎥
⎦
2𝑘
Resulting in 𝑁𝑃,𝑚𝑖𝑛 = 15. 74, as a result I go with 𝑁𝑃,𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑁2 = 16
To find max teeth of gear we use equation 3
2
3.) 𝑁
𝐺,𝑚𝑎𝑥
=
2
2
𝑁𝑃,𝑎𝑐𝑡𝑢𝑎𝑙 𝑠𝑖𝑛 (ϕ)−4𝑘
2
4𝑘−2𝑁𝑃,𝑎𝑐𝑡𝑢𝑎𝑙 𝑠𝑖𝑛 (ϕ)
Resulting in 𝑁𝐺,𝑚𝑎𝑥 = 101.07. I go with 𝑁𝐺,𝑎𝑐𝑡𝑢𝑎𝑙=𝑁3= 80 which satisfies equation 1
Now to find standard diametral pitch 𝑃23 we use equation
4.) 𝑃23 =
𝑁2+𝑁3
2𝑑𝑠
,resulting in 𝑃23= 8
The same goes for N4, N5 and P45
Using pressure angle ϕ = 20° and given center distance ds= 6 in
Now mG45 = 2
𝑁
𝑁
= 𝑁𝐺,𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑁5 = 2
𝐺,45
5.) 𝑚
𝑃,𝑎𝑐𝑡𝑢𝑎𝑙
4
Using equation 2 we get 𝑁𝑃,𝑚𝑖𝑛 = 14. 16 and go with 𝑁𝑃,𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑁4 = 16
Using equation 3 we get 𝑁𝐺,𝑚𝑎𝑥 = 101.07 and go with 𝑁𝐺,𝑎𝑐𝑡𝑢𝑎𝑙=𝑁5 = 32
Satisfying equation 4 for this instance
Then to find standard diametral pitch 𝑃45
𝑃45 =
𝑁4+𝑁5
2𝑑𝑠
, resulting in 𝑃45=4
Gear Analysis
For H = 12.5 hp, d3= 10 in, n = 100 rpm, β=20°, d2= 2in
𝑡,23
𝑊
𝑉 =
𝑟
𝐻
= 33000 𝑉 , =1575 lbf
π𝑑3𝑛
12
𝑡,23
𝑡𝑎𝑛β = 573. 48 lbf
𝐹 =𝑊
𝐹23 = 𝐹32 =
𝑡,23 𝑑3
2
𝑇𝑎3 =𝑊
𝑟 2
𝑡,23 2
(𝐹 ) + (𝑊
= 7875 lbf*in
) = 1676 lbf
𝑡,23 𝑑2
= 1575 lbf*in
2
𝑇𝑏3 = 𝑊
n=1000 rpm, H = 12.5 hp, β=20°,d4 = 4 in, d5 = 8 in
𝑡,45
𝑊
𝑉 =
𝑟
𝐻
= 33000 𝑉 , = 393.9 lbf
π𝑑4𝑛
12
𝑡,45
𝑡𝑎𝑛β = 143. 3 lbf
𝐹 =𝑊
𝐹45 = 𝐹54 =
𝑟 2
𝑡,45 2
(𝐹 ) + (𝑊
𝑡,45 𝑑4
2
= 786 lbf*in
𝑡,23 𝑑5
2
= 1572lbf*in
𝑇𝑐4 =𝑊
𝑇𝑑5 =𝑊
) = 418 lbf