The link Shown is 925 in. Thick, and has a single. Shear Pin connection. what is most nearly the required diameter (in) of the Pinif (tis average allowable shear Stress is 12ksi? ↓ 4 2in 20ksi

T'he pin is made of a material having a failure shear stress of Tail = 100 MPa. Determine the minimum required diameter of the pin to the nearest mm. Apply a factor of safety of F.S. = 2.5 against shear failure. %3D 80 kN

(2) 4 0 5 211 5 4 #9 4 "1 16 #3" ft-k fé = 3 ksi, M = 90 NWC. The section is cracked. Computer the bending stresses in the beam by using transformed-area metho

I need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala For the beam loaded as shown, find the maximum tensile and compressive bending stresses and maximum shearing stress.. IN.A=4*10mm* 80mm 100mm 60mm 60mm 1.5KN 2kN/m 1mal A |-----|- | Im ++ Im->> 3m 3kN/m