Answer Happy

Posted: **Mon May 09, 2022 10:16 am**

: 3 For W 150 Kn M A 50 Mm And A Pin Diameter Of 20 Mm The Average Shear Stress In The Pin Is Best Given By X 23 1 (18.59 KiB) Viewed 23 times

3. For w = 150 kN/m, a = 50 mm, and a pin diameter of 20 mm, the average shear stress in the pin is best given by X. 23.9 MPa b. 5.97 MPa 11.9 MPa d. 47.7 MPa r-w 150 A T202 A: 314,15 لے 2 pin ponts «43 ПРА 150 T202= 1250=11.9