Problem 2 to Problem 8 are all based on following problem st

Problem 2 to Problem 8 are all based on following problem statement In Fig. 2, the double angle 214x3x3/8 of A36 is subject to tensile load, among which dead load T 100 kips and live load TL 25 kips. The bolt diameter is 7/8 in. in standard hole. The strength of A36 steel is Fy-36 ksi and Fu-58 ksi. 3/4 Ag-248 in one ange) 2L4x3x3/8 X 0.782 Fig.2 Problem 2: What is design load T, that will use to check the double angle 2L4xs (show calculation and indicate one that will be used)? (15 points)

Solution

answer

given by

Given Data:

Used division is 2L4X3 1/2x1/2 , it is a american section. property are taken from AISC and these are provide below.

Total disgusting cross sectional area (Ag) = 7.01 in2 and disgusting area (ag) of a single angle is 3.5 in2;

subdivision Modulus about x bearing (Major axis) = 3.0 in3 and radious of twirl along x (Major axis) = 1.04 in ;

Yield vigor of steel 36 Ksi and Ultimate potency of steel 58 Ksi;

Solution:

For the single point of view,

The ostensible potency based on disgusting area (Pn) = Fy x ag = (36 x 3.5) = 126 Kips

at hand are three bolt in each angle so the net vicinity of one angle (anet) = 3.5 - 1/2x3x1 = 2 in2 , hole dia is 1in

The useful net area of the solo angle is (Ae) = \\Phit x anet = 0.75 x 2 = 1.5 in2

The nominal potency based on the grid area (Pn) = Fu x Ae = 58 x 1.5 = 87 Kips.

For the solitary angle,

The plan strength based on disgusting area (Pn) = 0.9 x 126 = 113.4 Kips

The Design power based on mesh area (Pn) = 0.75 x 87 = 65.25 Kips

So, the lesser worth should be in use for design i.e 65.25 Kips.

For double perspective the design delivery would be (Tu) = (65.25 x 2) = 130.5 Kips.