The tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and Fu= 400 MPa,and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD.40mm 40mn 40mm50mm...Compute the allowable strength based on yielding in KN.0Compute the nominal strength based on fracture in KN.0Check the adequacy of the section.0

Given : L150X90X10 A36 steel : Fy = 250 MPa, Fu = 400 MPa DL = 150kN, LL = 150 kN Bolts are 20 mm…

The tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and F,= 400 MPa, and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD. 50mm ... 60m Compute the allowable strength based on yielding in KN.0 Compute the nominal strength based on fracture in KN.O Check the adequacy of the section.0

The tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and F,= 400 MPa, and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD. 50mm ... 60m Compute the allowable strength based on yielding in KN.0 Compute the nominal strength based on fracture in KN.O Check the adequacy of the section.0

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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Question

The tension member shown is an angle L150 x 90 x 10, of A36 steel, with Fy= 250 MPa and Fu= 400 MPa,
and is subjected to a service DL= 150 KN and a service LL= 150 KN The bolts are 20 mm o. Use ASD.
40mm 40mn 40mm
50mm
...
Compute the allowable strength based on yielding in KN.0
Compute the nominal strength based on fracture in KN.0
Check the adequacy of the section.0

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