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. steelbolts evenly spaced around a bolt circle 12 in. indiameter, and six 4 -in. aluminum bolts on aconcentric bolt circle 8 in. in diameter. What torquecan be applied without exceeding 15000 psi in thesteel or 10000 psi in the aluminum? Assume Gst =12 x 106 psi and Gal = 4 x 106 psi. Problem 331A flanged bolt coupling consists of six ½ -in. steel bolts evenly spaced around a boltcircle 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 or6000 psi in the aluminum? Assume Gst = 12 x 10° psi and Ga = 4 x 10° psi.Solution 331T= (PRn),; + (PRn)aT= (AtRn)s: + (AtRn)aT=D 올짜42)2ta(6)(6) + 을 찌94)2ra(4) (4)T= 2.25nT + 2.25TT= 2.25T(Tst + Tai)→ Equation (1)GRGR(12x10°)(6)(4x 10°)(4)→ Equation (2a)Tal = Tst→ Equation (2b)Equations (1) and (2a)T-2.25π(을 Tai + Ta)=12.375TTalT = 12.375r(6000) = 74 250n lb-inT= 233.26 kip-inEquations (1) and (2b)T= 2.25r(Ts: + Ts:) = 2.75nts:T= 2.25t(9000) = 24 750n lb-inT= 77.75 kip-inUse T = 77.75 kip-in

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.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Connections

In steel structures, connections are the members that join two or more steel members. The connections are the most important members of any steel structure, and hence, they are designed more carefully. The majority of the construction cost of steel structures is incurred due to the connections. The design of connections is done as per the recommendations and specifications of the American Institute of Steel Construction (AISC).

Question

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.

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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