Please explain every detail to me on how they get the 45 that I encircled in a blue pen. Thank you so much! I'll rate your answer. KAX 12 +63X6 +83Y12=70X6+ suX 3ToknShen83kn.soen83kn16mAE3m B3m c3mEME=ORA = -77kdRE tRA = 70+so"E= 197knHA= 83+83=1G6 lenJomtohoc= 127.Skn|JointAJointEtmo = 6FAF83knO = 63,435¢166kn77kn86.089FAF Smo =77FAF = 86.089kgFGFCO10+fcf lO10 = 86.089 CO10-83fef Bimo - fef simo =86.029 Aimel.02) conofGF=-86089Fof =1797166= FA8+TFAB = 127, Stn top chadmamhenllenythfocef | tmit force (U|AB127.5O,5BC3.127.50.50.5CD98.5DE3.98.5O.5AFSus8.089otAtFGJus-6:708Sys– 220, 25| 71·1/BAtEHHGJys|-|92,302 -1.|18BFDH12108fC-१२.११7CHSus27.9S06ZGH = PUL) -127. SX0.SX3X2+ 98.5XO.S 3Y2 +AEAE86.89 Y 1.\18Y Sus -6,708X1.118X Sys t 220,2SX1.118XSTSt 192.30 ?X1:1/ 8X T4s)= 4367.385X100800X So04-Pay= 43167 mm -

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Answered: Please explain every detail to me on…

Please explain every detail to me on how they get the 45 that I encircled in a blue pen. Thank you so much! I'll rate your answer.

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

Introduction to Engineering Design

Students are often confused with engineering and design. The term engineering design has two words with different meanings, engineering, and design. The term engineering implies applications of laws and principles of science to design and analyze physical structures and components. In civil engineering, students apply the concept of physics and chemistry to design and build structures and buildings. In engineering disciplines, the term design means determining the dimensions, loads, and associated influences on the component due to the applied loads such as stresses and deformations. The estimation of such parameters is crucial to make the component or structure foolproof.

Question

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Please explain every detail to me on how they get the 45 that I encircled in a blue pen. Thank you so much! I'll rate your answer.

KAX 12 +63X6 +83Y12=70X6+ suX 3
Tokn
Shen
83kn.
soen
83kn
16m
A
E
3m B
3m c
3m
EME=O
RA = -77kd
RE tRA = 70+so
"E= 197kn
HA= 83+83=1G6 len
Jomto
hoc= 127.Skn|
JointA
JointE
tmo = 6
FAF
83kn
O = 63,435¢
166kn
77kn
86.089
FAF Smo =77
FAF = 86.089kg
FGFCO10+fcf lO10 = 86.089 CO10-83
fef Bimo - fef simo =
86.029 Aime
l.02) cono
fGF=-86089
Fof =1797
166= FA8+
TFAB = 127, Stn

top chad
mamhenl
lenyth
focef | tmit force (U|
AB
127.5
O,5
BC
3.
127.5
0.5
0.5
CD
98.5
DE
3.
98.5
O.5
AF
Sus
8.089
ot
At
FG
Jus
-6:708
Sys
– 220, 25| 71·1/B
At
EH
HG
Jys
|-|92,302 -1.|18
BF
DH
12
108
fC
-१२.११7
CH
Sus
27.9S06
ZGH = PUL) -127. SX0.SX3X2+ 98.5XO.S 3Y2 +
AE
AE
86.89 Y 1.\18Y Sus -6,708X1.118X Sys t 220,2SX1.118XSTS
t 192.30 ?X1:1/ 8X T4s)
= 4367.385X100
800X So0
4-
Pay= 43167 mm -

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