Instructions:-Write down each step clearly in calculation suggestion what you are doing.( you need to tell why you selected particular value ofø (phi) and U). What and why you are calculating certain stepConsider R= 433.5PROBLEMA welded tension member is consisting of two channels placed 400mm back to back with flanges turned out. Select a channel for factored tensile of 4RKN using A36 steel and AISC specifications.The member is to be 15m long.PLASE USE THE EXAMPLE AS GUIDE OF ALL STEPS AND FORMULAS TO BE USE 52STEEL STRUCTURESSIDDIQI Z. A.DESIGN OF TENSION MEMBERS $3CHAPTER 2:Solution:Approximate r, and r,: (using Design Aids)P, - 450 kNT,= 840 kN(1 +0.015 L) P.r,= 0.36 h = 0.36(152) - 55 mmr,-0.60 b - 0.60(300 – 2x51) - 118.8 mm- (1.54) (450)- 693 kNT, > (1 +0.015 L) P.Exact r, & r, (preferable and a must for final trial):152Referring to Figure 2.17,r, = 56.4 mm as for a single sectionI, =2 x 36.0 x10 + 2 x 1990 (150 - S1 + 12.7)- 5038 x 10' mm'Design first as a tension member and then check for P. (Case 3).For welded connections,3005038 x 10- 112.5 mmI, x1000Areg840x10002x 1990Figure 2.17. Built-Up Section MadeBy Two Channels.= 3733 mm0.9F,0.9x2503733Areg for one channel == 1867 mmPain 56.4 mm%3Dbain - 50 mn for welded connections6x 1000Li Tmin.106.4 < 200(OK)564Options available:1. C 150 x 15.62. МС 310 х 15.8Design of Lacing: This part can be completed after discussion on the design of compressionmembers.e.gTrial section: 2C, 150 x 15.6 (Figure 2.16)A -1990 mm?d =152mm ; by=5ImmI = 633 x 10ʻ manI, -36.0 x 10' mm- 12.7 mn, - 56.4 mm ; r,-13.4 mmCheck For Compressive Strength:** x 200,000(106.4)*F.174.36 kNFigure 2.16. Location of Centroidfor a Channel Section.(KL./r)> 0.44 F,- 110 kN(Inelastic buckling)Check for bte4F -F, = 0.9x p.658 * 250Both d and by > buin(OK)123.47 MPaCapacity Check:123.47 х 2х 1990/1000AT,49141 kNP.(OK)0.9 F, Aal=0.9 x 250 x (2 x 1990) / 1000AT. - 0.75 F,UA.895.5 kN > T. (OK)Loading cycles are assumed less than 20,000= 0.75 x 400 x 0,85 x (2x1990y1000 - 1014.9 kN > T,(OK)Assume U-0.85 from design aids and later check by using the detailed formula.Design Connections

Given data, Member details Grade of steel=A36 Span of member,L=15 m Load Factored design…

Answered: A welded tension member is consisting…

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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Instructions:-
Write down each step clearly in calculation suggestion what you are doing.( you need to tell why you selected particular value of
ø (phi) and U). What and why you are calculating certain step

Consider R= 433.5

PROBLEM

A welded tension member is consisting of two channels placed 400mm back to back with flanges turned out. Select a channel for factored tensile of 4RKN using A36 steel and AISC specifications.
The member is to be 15m long.

PLASE USE THE EXAMPLE AS GUIDE OF ALL STEPS AND FORMULAS TO BE USE

52
STEEL STRUCTURES
SIDDIQI Z. A.
DESIGN OF TENSION MEMBERS $3
CHAPTER 2:
Solution:
Approximate r, and r,: (using Design Aids)
P, - 450 kN
T,= 840 kN
(1 +0.015 L) P.
r,= 0.36 h = 0.36(152) - 55 mm
r,-0.60 b - 0.60(300 – 2x51) - 118.8 mm
- (1.54) (450)
- 693 kN
T, > (1 +0.015 L) P.
Exact r, & r, (preferable and a must for final trial):
152
Referring to Figure 2.17,
r, = 56.4 mm as for a single section
I, =2 x 36.0 x10 + 2 x 1990 (150 - S1 + 12.7)
- 5038 x 10' mm'
Design first as a tension member and then check for P. (Case 3).
For welded connections,
300
5038 x 10
- 112.5 mm
I, x1000
Areg
840x1000
2x 1990
Figure 2.17. Built-Up Section Made
By Two Channels.
= 3733 mm
0.9F,
0.9x250
3733
Areg for one channel =
= 1867 mm
Pain 56.4 mm
%3D
bain - 50 mn for welded connections
6x 1000
Li Tmin.
106.4 < 200
(OK)
564
Options available:
1. C 150 x 15.6
2. МС 310 х 15.8
Design of Lacing: This part can be completed after discussion on the design of compression
members.
e.g
Trial section: 2C, 150 x 15.6 (Figure 2.16)
A -1990 mm?
d =152mm ; by=5Imm
I = 633 x 10ʻ man
I, -36.0 x 10' mm
- 12.7 mn
, - 56.4 mm ; r,-13.4 mm
Check For Compressive Strength:
** x 200,000
(106.4)*
F.
174.36 kN
Figure 2.16. Location of Centroid
for a Channel Section.
(KL./r)
> 0.44 F,
- 110 kN
(Inelastic buckling)
Check for bte
4F -
F, = 0.9x p.658 * 250
Both d and by > buin
(OK)
123.47 MPa
Capacity Check:
123.47 х 2х 1990/1000
AT,
49141 kN
P.
(OK)
0.9 F, Aal
=0.9 x 250 x (2 x 1990) / 1000
AT. - 0.75 F,UA.
895.5 kN > T. (OK)
Loading cycles are assumed less than 20,000
= 0.75 x 400 x 0,85 x (2x1990y1000 - 1014.9 kN > T,
(OK)
Assume U-0.85 from design aids and later check by using the detailed formula.
Design Connections

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