PER IS 800:2007,Pmin = 2.5 × dn = 2.5 × 20 = 50 mmemin = 1.5 × dh = 1.5 × 22 = 33 mme = edge distanceP = PitchLet n be the no of bolts that can be ACCOMMODATED in a single row of width 140 mm. 2 × e + (n - 1) × P = 140To get maximum VALUE of n, e and P should be minimum i.e. e = emin and P = Pmin2 × 33 + (n - 1) × 50 = 140∴ n = 2.48 ∴ Maximum no of bolts that can be provided are 2.

"> PER IS 800:2007,Pmin = 2.5 × dn = 2.5 × 20 = 50 mmemin = 1.5 × dh = 1.5 × 22 = 33 mme = edge distanceP = PitchLet n be the no of bolts that can be ACCOMMODATED in a single row of width 140 mm. 2 × e + (n - 1) × P = 140To get maximum VALUE of n, e and P should be minimum i.e. e = emin and P = Pmin2 × 33 + (n - 1) × 50 = 140∴ n = 2.48 ∴ Maximum no of bolts that can be provided are 2.

">
Mobile Technologies Web Technologies Digital Marketing Creative Design Career Talk Medical General Tech Interviews Internet Of Things

Embark on a journey of knowledge! Take the quiz and earn valuable credits.

Take A Quiz

Challenge yourself and boost your learning! Start the quiz now to earn credits.

Take A Quiz

Unlock your potential! Begin the quiz, answer questions, and accumulate credits along the way.

Take A Quiz

Popular Categories

All
Interview Questions
QuestionBank
Quantitative Aptitude
Engineering
Govt Exams
Digital Marketing
Management
Interviews

What is the maximum number of 20 mm diameter bolts that can be accommodated in a single row on a 140 mm wide flat strip used as one of the structural elements involved in the process?

General Knowledge General Awareness in General Knowledge 9 months ago

  1.57K   0   0   0   0 tuteeHUB earn credit +10 pts
5 Star Rating 1 Rating

The relation between nominal and bolt hole diameter is given below:Nominal diameterHole size12 - 14 mm1 mm extra16 - 24 mm2 mm extra> 24 mm3 mm extra The nominal diameter of bolt, dn = 20 mm∴ Diameter of hole, dh = 20 + 2 = 22 mmAs PER IS 800:2007,Pmin = 2.5 × dn = 2.5 × 20 = 50 mmemin = 1.5 × dh = 1.5 × 22 = 33 mme = edge distanceP = PitchLet n be the no of bolts that can be ACCOMMODATED in a single row of width 140 mm. 2 × e + (n - 1) × P = 140To get maximum VALUE of n, e and P should be minimum i.e. e = emin and P = Pmin2 × 33 + (n - 1) × 50 = 140∴ n = 2.48 ∴ Maximum no of bolts that can be provided are 2.

Previous Next

Posted on 19 Nov 2024, this text provides information on General Knowledge related to General Awareness in General Knowledge. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

Take Quiz To Earn Credits!

Turn Your Knowledge into Earnings.

tuteehub_quiz

Tuteehub forum answer Answers

Post Answer

No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.

Explore Other Libraries

Blogs Tutorials Question Bank Feeds Careernews