Homework Help is Here – Start Your Trial Now!
Engineering
Civil Engineering
Use latest NSCP/NSCP 2015 formula to solve this problem *Use hand written to solve this problem *Kindly refer my picture attached  presented formula for block shear .   The butt connection shows 8-22 mm dia. A325 bolts spaced as follows: S1 = 40 mm S3 = 50 mm t1 = 16 mm S2 = 80 mm S4 = 100 mm t2 = 12 mm   Steel strength and stresses are: Fy = 248 MPa Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt Hole diameter = 25 mm   Questions:  A. Based on block shear strength

Use latest NSCP/NSCP 2015 formula to solve this problem *Use hand written to solve this problem *Kindly refer my picture attached  presented formula for block shear .   The butt connection shows 8-22 mm dia. A325 bolts spaced as follows: S1 = 40 mm S3 = 50 mm t1 = 16 mm S2 = 80 mm S4 = 100 mm t2 = 12 mm   Steel strength and stresses are: Fy = 248 MPa Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt Hole diameter = 25 mm   Questions:  A. Based on block shear strength

Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Charles Gilmore
Chapter8: Engineering Materials
Section: Chapter Questions
Problem 85CQ
See similar textbooks
icon
Related questions
Question

Topic:Bolted Steel Connection - Civil Engineering

 

*Use latest NSCP/NSCP 2015 formula to solve this problem

*Use hand written to solve this problem

*Kindly refer my picture attached  presented formula for block shear .

 

The butt connection shows 8-22 mm dia. A325 bolts spaced as follows:

S1 = 40 mm S3 = 50 mm t1 = 16 mm

S2 = 80 mm S4 = 100 mm t2 = 12 mm

 

Steel strength and stresses are:

Fy = 248 MPa Fu = 400 MPa

Allowable tensile stress on the gross area = 148 MPa

Allowable tensile stress on the net area = 200 MPa

Allowable shear stress on the net area = 120 MPa

Allowable bolt shear stress, Fv = 120 MPa

Bolt Hole diameter = 25 mm

 

Questions:

 A. Based on block shear strength

 

R₁ = 0.6FuAnv + Ups FuAnt ≤ 0.6FyAgv + Ubs FuAnt kindly, use this formula p=0.75(LRFD) where Agu Ant Anv n = 2.00 (ASD) gross area subject to shear, mm² net area subject to tension, mm² net area subject to shear, mm² Where the tension stress is uniform, Ubs = 1; where the tension stress is nonuniform, Ubs = 0.5.
Transcribed Image Text:R₁ = 0.6FuAnv + Ups FuAnt ≤ 0.6FyAgv + Ubs FuAnt kindly, use this formula p=0.75(LRFD) where Agu Ant Anv n = 2.00 (ASD) gross area subject to shear, mm² net area subject to tension, mm² net area subject to shear, mm² Where the tension stress is uniform, Ubs = 1; where the tension stress is nonuniform, Ubs = 0.5.
t1 T # Iss S3 S4 S3 # 53 54 53 t2 t2 →T S1 S2 S1
Transcribed Image Text:t1 T # Iss S3 S4 S3 # 53 54 53 t2 t2 →T S1 S2 S1
Expert Solution
steps

Step by step

Solved in 6 steps with 7 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Bars
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Materials Science And Engineering Properties
Materials Science And Engineering Properties
Civil Engineering
ISBN:
9781111988609
Author:
Charles Gilmore
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS