Hi,
A simple truss has pin-connected members. The members carry only axial forces, but the pins have to bear the shear stresses. Given example shows the calculation for the minimum required diameter of pin connected in double shear.
Problem: Two structural members AB and BC (shown below) are connected by a pin at B that is subjected to two 5-kip horizontal forces. The pins used at A and C are identical to the pin at B. The allowable shear stress for pins is 15 ksi, and the allowable bearing stress for members AB and BC is 20 ksi. Determine:
(a) the maximum average normal stress in members AB and BC.
(b) the minimum required diameters of the pins.
Solution:
A simple truss has pin-connected members. The members carry only axial forces, but the pins have to bear the shear stresses. Given example shows the calculation for the minimum required diameter of pin connected in double shear.
Problem: Two structural members AB and BC (shown below) are connected by a pin at B that is subjected to two 5-kip horizontal forces. The pins used at A and C are identical to the pin at B. The allowable shear stress for pins is 15 ksi, and the allowable bearing stress for members AB and BC is 20 ksi. Determine:
(a) the maximum average normal stress in members AB and BC.
(b) the minimum required diameters of the pins.
Solution:
No comments:
Post a Comment