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bolts evenly spaced around a bolt circle 12 in. in diameter, and six 4 -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 15000 psi in the steel or 10000 psi in the aluminum? Assume Gst = 12 x 106 psi and Gal = 4 x 106 psi.

bolts evenly spaced around a bolt circle 12 in. in diameter, and six 4 -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 15000 psi in the steel or 10000 psi in the aluminum? Assume Gst = 12 x 106 psi and Gal = 4 x 106 psi.

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter7: Simple Connections
Section: Chapter Questions
Problem 7.11.10P
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Hi, I have a question and i also attached a similar problem but with different number of coupling bolts for your solving reference. My question has eight coupling bolts in it. Thank you.

15. A flanged bolt coupling consists of eight ½ -in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and six 4 -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 15000 psi in the steel or 10000 psi in the aluminum? Assume Gst = 12 x 106 psi and Gal = 4 x 106 psi.
Transcribed Image Text:15. A flanged bolt coupling consists of eight ½ -in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and six 4 -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 15000 psi in the steel or 10000 psi in the aluminum? Assume Gst = 12 x 106 psi and Gal = 4 x 106 psi.
Problem 331 A flanged bolt coupling consists of six ½ -in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and four ¾a -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 9000 psi in the steel or 6000 psi in the aluminum? Assume Gst = 12 x 10° psi and Ga = 4 x 10° psi. Solution 331 T= (PRn),; + (PRn)a T= (AtRn)s: + (AtRn)a T=D 올짜42)2ta(6)(6) + 을 찌94)2ra(4) (4) T= 2.25nT + 2.25T T= 2.25T(Tst + Tai) → Equation (1) GR GR (12x10°)(6) (4x 10°)(4) → Equation (2a) Tal = Tst → Equation (2b) Equations (1) and (2a) T-2.25π(을 Tai + Ta)=12.375TTal T = 12.375r(6000) = 74 250n lb-in T= 233.26 kip-in Equations (1) and (2b) T= 2.25r(Ts: + Ts:) = 2.75nts: T= 2.25t(9000) = 24 750n lb-in T= 77.75 kip-in Use T = 77.75 kip-in
Transcribed Image Text:Problem 331 A flanged bolt coupling consists of six ½ -in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and four ¾a -in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied without exceeding 9000 psi in the steel or 6000 psi in the aluminum? Assume Gst = 12 x 10° psi and Ga = 4 x 10° psi. Solution 331 T= (PRn),; + (PRn)a T= (AtRn)s: + (AtRn)a T=D 올짜42)2ta(6)(6) + 을 찌94)2ra(4) (4) T= 2.25nT + 2.25T T= 2.25T(Tst + Tai) → Equation (1) GR GR (12x10°)(6) (4x 10°)(4) → Equation (2a) Tal = Tst → Equation (2b) Equations (1) and (2a) T-2.25π(을 Tai + Ta)=12.375TTal T = 12.375r(6000) = 74 250n lb-in T= 233.26 kip-in Equations (1) and (2b) T= 2.25r(Ts: + Ts:) = 2.75nts: T= 2.25t(9000) = 24 750n lb-in T= 77.75 kip-in Use T = 77.75 kip-in
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