Question

The rectangular bar is connected to the support bracket with a 10-mm-diameter pin. The bar width is w = 70 mm and the bar thickness is 20 mm. Each side of the bracket has the same dimensions as the bar. The average shear stress in the pin cannot exceed 115 MPa, the bearing stress in the bar cannot exceed 120 MPa, and the bearing stress in the bracket cannot exceed 120 MPa. Determine the maximum value of Pmax that can be supported by the structure.

Answer 1

Answer:

P_max = 9.032 KN

Step-by-step explanation:

Given:

- Bar width and each side of bracket w = 70 mm

- Bar thickness and each side of bracket t = 20 mm

- Pin diameter d = 10 mm

- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa

- Average allowable shear stress of pin S = 115 MPa

Find:

The maximum force P that the structure can support.

Solution:

- Bearing Stress in bar:

                                       T = P / A

                                       P = T*A

                                       P = (120) * (0.07*0.02)

                                       P = 168 KN

- Shear stress in pin:

                                        S = P / A

                                        P = S*A

                                        P = (115)*pi*(0.01)^2 / 4

                                        P = 9.032 KN

- Bearing Stress in each bracket:

                                       T = P / 2*A

                                       P = T*A*2

                                       P = 2*(120) * (0.07*0.02)

                                       P = 336 KN

- The maximum force P that this structure can support:

                                      P_max = min (168 , 9.032 , 336)

                                      P_max = 9.032 KN

Answer 2

In this exercise we have to use the knowledge of force to calculate the maximum force of P, in this way we can say that:

$P_{max} = 9.032 KN$

From the information given in the statement as:

• Bar width: $w = 70 mm$

• Bar thickness: $t = 20 mm$

• Pin diameter: $d = 10 mm$

• Average allowable bearing: $T = 120 MPa$

• Average allowable shear: $S = 115 MPa$

So calculating the bearing stress bar:

$T = P / A\\ P = T*A\\ P = (120) * (0.07*0.02)\\P = 168 KN$                                      

So calculating the shear stress:

                                     

 $S = P / A\\ P = S*A\\P = (115)*pi*(0.01)^2 / 4\\ P = 9.032 KN$

So calculating the Bearing Stress:

$T = P / 2*A\\ P = T*A*2\\ P = 2*(120) * (0.07*0.02)\\P = 336 KN$

So calculating the maximum force P that this structure can support:

$P_{max} = min (168 , 9.032 , 336)\\P_{max} = 9.032 KN$

See more about force at brainly.com/question/26115859