Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Question
Determine reactions AY, BY, AX, BX. Calculate normal force, shear force and bending moments for diagrams and draw it.
All needed informations:
Given:
L= 7meter
H1= 1meter
H2= 3meter
a= 1meter
b= 3meter
F1= -10kN
F2= 6kN
P1= -3kN/meter
P2= 2kN/meter
Transcribed Image Text:P₂
F₂₁
L
b
LL~
F₂
✓²
P₁
H1
*
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Follow-up Question
Is this right because the answers is different?
Transcribed Image Text:2K
Ay
-Ax
10
उन
16
1.5 1.5 1.5 1.5
74.
Br
3E
Bending moment at Hinge = 0
>-(Ayx 4) + (A₂x4) = (2x4) 1 - (10x1.5)
and, EMB: 0
> - (Ayx 7) + (A₂x¹) = -(2x4) (1-1)-(10×4.5) + 6× 1·5)
..Ay = 5.125 KN, A₂ = 5.375 km
Fy=0 ⇒
5.125+ 6-10 = 1.125 KN =
EF=0> B₂ = (3x3) + 5.375-(2x4) - 6.375 KN.
Solution
by Bartleby Expert
Follow-up Question
Can you please simlify that how you solve these equations to got ay and ax, step by step, thanks. (ay 5.125 kN and AX 5.375 kN)
Solution
by Bartleby Expert
Follow-up Question
This is different task, but can u show how to solve like this way?
Transcribed Image Text:Ax, Ay:
➡
→
Bx, By:
➡
Mnv - Ay 4 m + Ax
EMB=- Ay 6 m + Ax
Ay 14 m 90 kNm
Ay = +6,43 kN (1)
3 m + 30 kN
1 m - 30 kN
Mn,v - Ay 4 m + Ax
3 m + 75 kNm = 0
EMB= -Ay · 18 m + Ax 3 m + 165 kNm = 0
Mn,o= + By 2 m + Bx
ZMA + By
6 m - Bx
1,5 m +20 kN
1,5 m = 0
0,5 m +20 kN - 3,5 m = 0
sij. Mnv 6,434 m + Ax 3 m + 75 kNm = 0
Ax= -16,43 kN (+)
10 kN/m
Mn,o= + By
2 m + Bx 2 m = 0
EMA= - By
12 m + Bx2m + 190 kNm = 0
By 14 m
190 kNm
By = + 13,57 kN (1)
sij. Mn,o 13,57 kN 2 m + Bx 2 m = 0
Bx=13,57 kN (+)
2 m = 0
1 m - 20 kN 2,5 m - 30 kN - 1,5 m = 0
TAY
1m
*
30 kN
·3
Ax
TAY
• (-2)
1m
1,5 m
10 kN/m C
α
20 KN
1,5 m
20 KN
1,5 m 1,5 m
*
20 KN
E
Ax= 16,43 kN
Bx
2 m
2 m
Bx = 13,57 kN
*
By = 13,57 kN
1m 2m
Solution
by Bartleby Expert
Follow-up Question
Is there any simpler way to got AY,AX,BY,BX?
Solution
by Bartleby Expert
Follow-up Question
Is this way simplier to got AY, BY, AX,BX find out? Can u simplier that equations, i don't got 1.88kN and 0.38 kN. I only got 1.77 and 2 kN
Thanks
Transcribed Image Text:ΣΜΑ =
From beam ACDE,
0
0 = Ex4 + Ex4 + (−10) ×2.5 +1.88×-
X
9.02 = 4 Ex + 4 Ey
ΣΜΒ = 0
From beam EFB,
E
E =
0 = -Ex3 + Ex3 - 6x1.5 - · (-3) × ²32²2
(−3) ×-
-4.5 = - 3 Ex
3 E + 3 E
y
=
4.123²
2
...(1)
Solving eqn 1 and 2,
1.88 KN (→)
0.38 KN (↓)
..(2)
Solution
by Bartleby Expert
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Is this right because the answers is different?
Transcribed Image Text:2K
Ay
-Ax
10
उन
16
1.5 1.5 1.5 1.5
74.
Br
3E
Bending moment at Hinge = 0
>-(Ayx 4) + (A₂x4) = (2x4) 1 - (10x1.5)
and, EMB: 0
> - (Ayx 7) + (A₂x¹) = -(2x4) (1-1)-(10×4.5) + 6× 1·5)
..Ay = 5.125 KN, A₂ = 5.375 km
Fy=0 ⇒
5.125+ 6-10 = 1.125 KN =
EF=0> B₂ = (3x3) + 5.375-(2x4) - 6.375 KN.
Solution
by Bartleby Expert
Follow-up Question
Can you please simlify that how you solve these equations to got ay and ax, step by step, thanks. (ay 5.125 kN and AX 5.375 kN)
Solution
by Bartleby Expert
Follow-up Question
This is different task, but can u show how to solve like this way?
Transcribed Image Text:Ax, Ay:
➡
→
Bx, By:
➡
Mnv - Ay 4 m + Ax
EMB=- Ay 6 m + Ax
Ay 14 m 90 kNm
Ay = +6,43 kN (1)
3 m + 30 kN
1 m - 30 kN
Mn,v - Ay 4 m + Ax
3 m + 75 kNm = 0
EMB= -Ay · 18 m + Ax 3 m + 165 kNm = 0
Mn,o= + By 2 m + Bx
ZMA + By
6 m - Bx
1,5 m +20 kN
1,5 m = 0
0,5 m +20 kN - 3,5 m = 0
sij. Mnv 6,434 m + Ax 3 m + 75 kNm = 0
Ax= -16,43 kN (+)
10 kN/m
Mn,o= + By
2 m + Bx 2 m = 0
EMA= - By
12 m + Bx2m + 190 kNm = 0
By 14 m
190 kNm
By = + 13,57 kN (1)
sij. Mn,o 13,57 kN 2 m + Bx 2 m = 0
Bx=13,57 kN (+)
2 m = 0
1 m - 20 kN 2,5 m - 30 kN - 1,5 m = 0
TAY
1m
*
30 kN
·3
Ax
TAY
• (-2)
1m
1,5 m
10 kN/m C
α
20 KN
1,5 m
20 KN
1,5 m 1,5 m
*
20 KN
E
Ax= 16,43 kN
Bx
2 m
2 m
Bx = 13,57 kN
*
By = 13,57 kN
1m 2m
Solution
by Bartleby Expert
Follow-up Question
Is there any simpler way to got AY,AX,BY,BX?
Solution
by Bartleby Expert
Follow-up Question
Is this way simplier to got AY, BY, AX,BX find out? Can u simplier that equations, i don't got 1.88kN and 0.38 kN. I only got 1.77 and 2 kN
Thanks
Transcribed Image Text:ΣΜΑ =
From beam ACDE,
0
0 = Ex4 + Ex4 + (−10) ×2.5 +1.88×-
X
9.02 = 4 Ex + 4 Ey
ΣΜΒ = 0
From beam EFB,
E
E =
0 = -Ex3 + Ex3 - 6x1.5 - · (-3) × ²32²2
(−3) ×-
-4.5 = - 3 Ex
3 E + 3 E
y
=
4.123²
2
...(1)
Solving eqn 1 and 2,
1.88 KN (→)
0.38 KN (↓)
..(2)
Solution
by Bartleby Expert
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