Please enlighten me on how they get those values that I encircled in a red pen and I also input the full solution. Thank you so much! I'll vote (thumbs up) for 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 JointE83kn一。fCf86.089tafcOno+fcf (O10 =86.089 COAO- 83fef Bimo - fef sino86.039 ineFGF=-867089J4.797%3D

One can also use calculator to solve the mentioned two equations. It is two unknown with two…

Answered: Then Tokn 83kn. sokn 83kn 16m 3m 3m c

Engineering

Civil Engineering

Then Tokn 83kn. sokn 83kn 16m 3m 3m c

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

100%

Please enlighten me on how they get those values that I encircled in a red pen and I also input the full solution. Thank you so much! I'll vote (thumbs up) for 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

JointE
83kn
一。
fCf
86.089
tafcOno+fcf (O10 =
86.089 COAO- 83
fef Bimo - fef sino
86.039 ine
FGF=-867089
J4.797
%3D

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