Joint A FAB sin o=70 FAB = 218.75 KN (C) Joint B e20RN fBc = 218.45-6.4 218.719 FBC 80 = 212.35 KN(C) FABCOs > FAJ FAJ =207.8 kN (T) FBJ 19RN (C) Joint J 1990-0 90-e Joint C by substitution method fje=30.08 RNCT) FJK =17.66 RN(T) 2078 Joint K 17.16 FKLXO.6 →FKLXO.8=22-84 6.4 19 , K 17766 2288 177-64 212. 85E FDE =170.95RN(C) FKL X0.6 = 17.16 FKL= 28-6 RN 28.5 30.08 FKN = 131.9 kN fcp =177.35RNCC) %3D FCK = 28.6 KN ((C) FCL = B0.08 RNCT) Q. 8.) ZH=0 =DHA=0 ZV=0 Re+RI = 10+20 +20+20+20 +20 +20+20 t ZMA=0 20x3 +2OX6 +20x9 +20x12. +20 X 1s +20x18+20x21 +10X 24 = RqX2Y RI=8ORN() RA = 80RN(↑) 20RN 20RN 20KN 20RN ER 20kN 20PN Vo 1705 EO 17 구.35 20kN 1ORN 55.7 LORN G 212.35CT) H 218.7SIT) 3008 2213 28 286 C) 207.8 194-4 K +319(T) R ST). LT) tane = W12 RA e - tan- (0.33) = 18. 43° sin 9 = 032 cosO = O.95 7.66 207.8LT) 4EQ = AE AE COsO= AS AP = 3m sin 20= 0.6 coo 29 =0.8 EQ = 3-16m 24m 9-434 80-08 28.6

Joint A FAB sin o=70 FAB = 218.75 KN (C) Joint B e20RN fBc = 218.45-6.4 218.719 FBC 80 = 212.35 KN(C) FABCOs > FAJ FAJ =207.8 kN (T) FBJ 19RN (C) Joint J 1990-0 90-e Joint C by substitution method fje=30.08 RNCT) FJK =17.66 RN(T) 2078 Joint K 17.16 FKLXO.6 →FKLXO.8=22-84 6.4 19 , K 17766 2288 177-64 212. 85E FDE =170.95RN(C) FKL X0.6 = 17.16 FKL= 28-6 RN 28.5 30.08 FKN = 131.9 kN fcp =177.35RNCC) %3D FCK = 28.6 KN ((C) FCL = B0.08 RNCT) Q. 8.) ZH=0 =DHA=0 ZV=0 Re+RI = 10+20 +20+20+20 +20 +20+20 t ZMA=0 20x3 +2OX6 +20x9 +20x12. +20 X 1s +20x18+20x21 +10X 24 = RqX2Y RI=8ORN() RA = 80RN(↑) 20RN 20RN 20KN 20RN ER 20kN 20PN Vo 1705 EO 17 구.35 20kN 1ORN 55.7 LORN G 212.35CT) H 218.7SIT) 3008 2213 28 286 C) 207.8 194-4 K +319(T) R ST). LT) tane = W12 RA e - tan- (0.33) = 18. 43° sin 9 = 032 cosO = O.95 7.66 207.8LT) 4EQ = AE AE COsO= AS AP = 3m sin 20= 0.6 coo 29 =0.8 EQ = 3-16m 24m 9-434 80-08 28.6

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

Classification of Legal Structure

The first step for the establishment of any company is to find a suitable legal structure. To determine the appropriate legal structure for a specific company or a venture, professional guidance is necessary for the deep understanding of various structure classifications, long-term targets, and certain choices. The categorization of different business legal structures comprises liability implications, income and taxation considerations for the company owners.

Question

I need this answer handwritten mention question number also and please on white page I'll UPVOTE for sure but pls diagram neat and clean and dont worry it's not again honor code etc

Joint A
FAB sin o=70
FAB = 218.75 KN (C)
Joint B
e20RN
fBc = 218.45-6.4
218.719
FBC
80
= 212.35 KN(C)
FABCOs
> FAJ
FAJ =207.8 kN (T)
FBJ
19RN (C)
Joint J
1990-0
90-e
Joint C
by substitution method
fje=30.08 RNCT)
FJK =17.66 RN(T)
2078
Joint K
17.16
FKLXO.6
→FKLXO.8=22-84
6.4 19 , K
17766
2288
177-64
212. 85E
FDE =170.95RN(C)
FKL X0.6 = 17.16
FKL= 28-6 RN
28.5 30.08
FKN = 131.9 kN
fcp =177.35RNCC)
%3D
FCK
= 28.6 KN ((C)
FCL = B0.08 RNCT)

Q. 8.)
ZH=0 =DHA=0
ZV=0
Re+RI = 10+20 +20+20+20 +20 +20+20 t
ZMA=0
20x3 +2OX6 +20x9 +20x12. +20 X 1s +20x18+20x21 +10X 24 = RqX2Y
RI=8ORN()
RA = 80RN(↑)
20RN
20RN
20KN
20RN
ER
20kN
20PN
Vo 1705
EO
17 구.35
20kN
1ORN
55.7
LORN
G 212.35CT) H 218.7SIT)
3008
2213
28
286 C)
207.8
194-4 K
+319(T)
R
ST).
LT)
tane = W12
RA
e - tan- (0.33) = 18. 43°
sin 9 = 032
cosO = O.95
7.66
207.8LT)
4EQ = AE
AE COsO= AS
AP = 3m
sin 20= 0.6
coo 29 =0.8
EQ = 3-16m
24m
9-434
80-08
28.6

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