Problem 1 In the formula for calculating the block shear str

Problem 1: In the formula for calculating the block shear strength of bolt connection, i.e., Eq 3.9.4 on the text book, it contains 0.6F What is the physical meaning of 0.6F.? (10 points) Problem 2 to Problem 8 are all hased on following problem statement In Fig. 2, the double angle 21.4 x 3 ½ x1/2 of A36 is subject to tensile load, among which dead load T 120 kips and live load T-50 kips. The diameter of bolt hole is 1 in. in standard hole. The strength of A36 steel is Fy-36 ksi and Fu-58 ksi. 1 3/4 1.51 2.5 25 Age35 ir One angle) a4x3 ½ x1/2 X-1 Fig.2 Problem 2: What is design load T, that will use to check the double angle 214x3 ½ x1/2 strength (show calculation and indicate one that will be used)? (15 points)

Solution

1. Physical meaning of 0.6Fu is that,

shear stress for rupture of steel = 0.6 times the ultimate stress of steel.

It means that a shear stress of magnitude 0.6 times ultimate stress can cause rupture of stress in normal tension. This value is obtained from Von Mises criterion.

Same way you can also have, shear strength to cause yielding of steel is approximately 0.6 times (exactly 0.577) the yield strength of steel in simple tension.

So physically it represnts ultimate shear strength of steel.

2. As per AISC, following are load combinations

The design load should be maximum of above = 224 kips