ASSIGNMENT # 5_ ALTERNATIVES
Question1
Capital t
40 KN
20 KN
T
Cy
Cx
installment payments on your 5
2 . 5
5/3
10
Sent out loads happen to be converted into centered (point) loads first: 8(KN m) × 5(m) sama dengan 40 ( KN )
1
× 8(KN m) × 5(m) = 20 ( KN )
2
(Rectangular load)
(Triangular load)
And now, the equilibrium from the F. B. D:
+∑ C sama dengan 0:
M
+
40 × 2 . 5 + T × 5  20 × 5 five 3 + T bad thing 30 ×10 = 0 ⇒
→ ∑ times = 0: Cx  Tcos30 = 0 ⇒
F
+ ↑ Fy = zero:
∑
C y  40 & 23. thirtythree  twenty + twenty three. 33 desprovisto 30 = 0 ⇒
T sama dengan 23. 33 ( KN )
Cx = twenty. 20 ( KN )
C y = twentyfive ( KN )
PROJECT # 5_ SOLUTIONS
Ay
Question2
Ax
Before splitting up the body, get the exterior reaction pieces: MF=0
Ax(138. 56) – 25(120) = 0 � Ax = 21. 65N
FX=0 � Fx = 21. 65N
FY=0 � Ay sama dengan 25N
Consider body VOTRE:
MD=0
Cy(40) – 25(60) = 0 � Cy = 37. 5N
Fy=0
Dy – 25 – 37. your five � Dy = sixty two. 5N
Consider body HUB PAGES:
MB=0
21. 65(69. 282) – Cx(34. 64) – Cy(20) = 0 � Cx = 21. 65N
Fy=0
Cy – By simply = zero � By = 37. 5N
Fx=0
Bx  Cx – +21. sixtyfive = zero � Bx = zero
Fx
twentyone. 65
By simply
25
Bx
Dx
Dy
25
Dy
Cx
Dx
Cy
By simply
Bx
Consider body CE again:
Fx=0
Cx – Dx sama dengan 0 � Dx = 21. 65N
34. 64
Cx
34. 922
twentyone. 65
Cy
ASSIGNMENT # 5_ ALTERNATIVES
Question several
Because of proportion we analyze
one of the two hinged users, the
upper part is usually chosen as well as its F. N. D.
along with that for the connection in
D is usually drawn. As a result of symmetry,
the forces by S and A do not x–
elements, the two–force members
BD and COMPACT DISK exert causes of equivalent
magnitude B = C on the connection
at Deb. equilibrium of the connection
gives:
FX=0;
W cosθ & C cosθ – T =0 � 2B cosθ = Capital t �
B= T/(2 cosθ)
From the N. B. Deb. of the uppr part we all express the equilibrium of moments regarding point A. substituting S=800 N, as well as the expression for B provides:
MA=0;
To
T
To
(cos θ )(50) &
(sin θ )(36)  36(800)  (26) sama dengan 0
two cos θ
2 cos θ
two
Substituting (sin θ/cosθ = tanθ = 5/12) and solving pertaining to T give:
T (25 +
5(36)
 13) = 28, 800 �
2(12)
T=1477 (N)
Finally, equilibrium in the y–direction gives us:
FY=0;
800 
S – B. sinθ – A = 0
1477 your five
A = 0 �
2(12 as well as 13) 13
A = 492 (N)
ASSIGNMENT # 5_ ALTERNATIVES
Question 4
a) Bring FBD while shown and solve the reaction forces, we now have: M B 0 (12)(8)(4) (20)(2. 5) ( A cos 18)(12. 5) ( A trouble 18)(4) 0 A 25. 449kN
F
F
X
0 (25. 449)(sin 18) 20 BX 0 BX 27. 864kN
Y
0 (25. 449)(cos 18) (12)(8) BY 0 BY 71. 797kN
b) Minimize the light beam at E and solve part D?GGE, we have:
Meters E (25. 449)(cos 18)(3) 72. 610kN m
FE, shearforce (25. 449)(cos 18) 24. 203kN
FE, axialforce (25. 449)(sin 18) 7. 8642kN
c) This is a new issue as demonstrated. The introduction of the midspan (intermediate) hinge for C plus the pin by A creates a completely new FBD.
Cy
12x8
Cx
Ax
Cx
Ay
Cy
Bx
By
Consider body ALTERNATING CURRENT:
MA=0
� (12x8)(8. 5)+Cy(12. 5) = 0
Fy=0
� (12x8)+Cy +Ay sama dengan 0
� Cy= 66. 28 kN
� Ay= 30. 72 kN
Consider body BC:
MB=0
� (20)(2. 5)+Cx(4) = zero
Fx=0
� 20 Cx Bx sama dengan 0
Fy=0
� ByCy=0
� Cx= 12. 5 kN
� Bx= several. 5 kN
� By= 65. 28 kN
Consider body ALTERNATING CURRENT again:
Fx=0
� CxAx = zero
� Ax= 12. a few kN
You can examine the reliability of your remedy by taking into consideration the FBD with the entire body ACB as you would in part a
ASSIGNMENT # 5_ SOLUTIONS
Question a few
First consider the exterior equilibrium of
the body just before splitting up:
MA=0
Ecos30(0. 6) + Esin30(0. 7608)240(0. 4608)120(0. 5608)=0
� Electronic = 197. 65N
Fx=0
Ecos30 & Ax=0
� Ax = 171. 17N
Fy=0
Esin30 + Ay – 240  120=0
Ax
� Ay = 261. 18N
240N
120N
E
zero. 1608m
Ay
It is noted that BD is a 2force member, therefore the FBD is as shown. Consider physique CE:
MC=0
Esin30(0. 6)240(0. 3)120(0. 4) + BDsin38. 26(0. 3)=0
� BD = 326. 78N
Fx=0
Ecos30 + BDcos38. 21 + Cx=0
� Cx = 85. 42N
Fy=0 Cy=58. 82 kN...

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