Givon:- DAB = 110 MPa 6AC = 120 MPa (A) AB (A) AC = 800mmL = 400mm² B 40° A maximum safe value of w, w = ? 50° с

Structural Analysis
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Chapter2: Loads On Structures
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In step 3, where did you get the 0.777865? How did you solved it?
46 6:28 Q
Given:-
DAB = 110 mpa
6AC 120 MPa
(A) AB
(A) AL
Step2
b)
Now
800mmL
= 400mm²
fx = 0
-FAB (0840+ FAC C0860 = 0
{fy=0
Step3
maximum safe value of w, W = ?
FAB CO340 = FAC (O360°
FAB =
FAC X (0860"
(0840
FAB = 0.6527 FAC
B
|||
(1)
FAB
40⁰°
FAB Singo + FAC Singo = w
D
A
W
W
(2)
60°
FAC
√x
U
Do
Transcribed Image Text:46 6:28 Q Given:- DAB = 110 mpa 6AC 120 MPa (A) AB (A) AL Step2 b) Now 800mmL = 400mm² fx = 0 -FAB (0840+ FAC C0860 = 0 {fy=0 Step3 maximum safe value of w, W = ? FAB CO340 = FAC (O360° FAB = FAC X (0860" (0840 FAB = 0.6527 FAC B ||| (1) FAB 40⁰° FAB Singo + FAC Singo = w D A W W (2) 60° FAC √x U Do
46 6:28 Q
4G
↑
Step3
from (1) 8 (11
0.6527 FAC X Singo + FAC XSin 6⁰ = w
1.28 557 FAC = W
FAC = 0.777865 W
from
FAB = 0.6527 FAC
|||
FAB= 0. 5077 W
=
= 0.6527 X 0.77786S W
#gf AB fails then
FAB 110 MPa = FAB 0.5077 W
(A)AB
800
110 = 0·5077 W
800
#gf As fails then
бас = 120 = гас
(A)AC
a
W = 1733 30.71 N
W 173.33 KN
-
120 = 0.7778654
400
W = 61707-37 N
W = 61-70 KN
F maximum safe value of W =
=
?
(1
["FAB = 0·6527 FAC]
0.777865 W
400
(OK L)
61.70 KN
√x
U
46
An
Do
Transcribed Image Text:46 6:28 Q 4G ↑ Step3 from (1) 8 (11 0.6527 FAC X Singo + FAC XSin 6⁰ = w 1.28 557 FAC = W FAC = 0.777865 W from FAB = 0.6527 FAC ||| FAB= 0. 5077 W = = 0.6527 X 0.77786S W #gf AB fails then FAB 110 MPa = FAB 0.5077 W (A)AB 800 110 = 0·5077 W 800 #gf As fails then бас = 120 = гас (A)AC a W = 1733 30.71 N W 173.33 KN - 120 = 0.7778654 400 W = 61707-37 N W = 61-70 KN F maximum safe value of W = = ? (1 ["FAB = 0·6527 FAC] 0.777865 W 400 (OK L) 61.70 KN √x U 46 An Do
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