Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
(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
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
See similar textbooks
icon
Related questions
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)
Transcribed Image Text: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)
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Stress
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS