Please explain why we divided by 2, and how we knew it is double shear PI30 Rigid bar ABC shown in Figure Pl.30 issupported by a pin at bracket Aand by tie rod (1).Tie rod (1) has a diameter of 5 mm, and it issupported by double-shear pin connections at BandD. The pin at bracket Ais a single-shearconnection. All pins are 7 mm in diameter.Assume a - 600 mm, b-300 mm, h-450 mm, P- 900 N, and e= 55. Determine the following:(a) the normal stress in rod (1)(b) the shear stress in pin B(c) the shear stress in pin AFIGURE PL30SolutionEquilibrium: Using the FBD shown, calculatethe reaction forces that act on rigid bar ABC.EM, = F, sin(36.87°)(600 mm)-(00 N)sin (55 X900 mm) =0F = L843.092 NEF,=A-(1,843.092 N)cos(36.87)+(900 N)cos(55) 0A =958.255 NEF, A+(1.843.092 Njsin(36.87)-(900 N)sin(55) 0A--368.618 NThe resultant force at Ais14-J958.255 N) +(-368.618 Ny =1,026.709 N(a) Normal stress in rod (1).A-(5 mm) = 19.635 mm1,843.092 N19.635 mm93.9 MPaAs.(b) Shear stress in pin B. The cross-sectional arca of a 7-mm-diameter pin is:4-7 mm) 38.485 mmPin Bis a double shear connection; therefore, its average shear stress isPlease explain why we divided by 2,and how we knew it is double shear1,843.092 N23.9 MPa2(38.485 mm)

Here, we will see 1.What is double shear 2.The effect of double shear 3.Its significance in given…

(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is: 4-7 mm' 38.485 mm Pin Bis a double shear connection; therefore, its average shear stress is 1,843.092 N Please explain why we divided by 2, and how we knew it is double shear 2(38.485 mm) 23.9 MPa

(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is: 4-7 mm' 38.485 mm Pin Bis a double shear connection; therefore, its average shear stress is 1,843.092 N Please explain why we divided by 2, and how we knew it is double shear 2(38.485 mm) 23.9 MPa

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Please explain why we divided by 2, and how we knew it is double shear

PI30 Rigid bar ABC shown in Figure Pl.30 is
supported by a pin at bracket Aand by tie rod (1).
Tie rod (1) has a diameter of 5 mm, and it is
supported by double-shear pin connections at Band
D. The pin at bracket Ais a single-shear
connection. All pins are 7 mm in diameter.
Assume a - 600 mm, b-300 mm, h-450 mm, P
- 900 N, and e= 55. Determine the following:
(a) the normal stress in rod (1)
(b) the shear stress in pin B
(c) the shear stress in pin A
FIGURE PL30
Solution
Equilibrium: Using the FBD shown, calculate
the reaction forces that act on rigid bar ABC.
EM, = F, sin(36.87°)(600 mm)
-(00 N)sin (55 X900 mm) =0
F = L843.092 N
EF,=A-(1,843.092 N)cos(36.87)+(900 N)cos(55) 0
A =958.255 N
EF, A+(1.843.092 Njsin(36.87)-(900 N)sin(55) 0
A--368.618 N
The resultant force at Ais
14-J958.255 N) +(-368.618 Ny =1,026.709 N
(a) Normal stress in rod (1).
A-(5 mm) = 19.635 mm
1,843.092 N
19.635 mm
93.9 MPa
As.
(b) Shear stress in pin B. The cross-sectional arca of a 7-mm-diameter pin is:
4-7 mm) 38.485 mm
Pin Bis a double shear connection; therefore, its average shear stress is
Please explain why we divided by 2,
and how we knew it is double shear
1,843.092 N
23.9 MPa
2(38.485 mm)

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