Solved problems for net area estimation.
How do we estimate the net area for plates?
Problem 3-1.
In problem 3-1, we have a plate 3/8 inch thick and 8 inches wide.
Determine the net area of the 3/8-inch x 8-inch plate, which is connected at its end with two lines of 3/4-inch bolts.
This is the original plate of 8 inches by 3/8 inches connected with two thin plates. Each has the same width of 8 inches and thickness of 1/4 inch.
This is the first gauge line in the direction of the longitudinal tension, which is in the direction of the applied force, and this is the second gauge line.
As discussed earlier, the distance between the two center lines at the longitudinal direction represents the gage distance, while the perpendicular distance is the pitch.
We don’t have a zigzag line, so we will deal with the vertical section, as we can see in the next slide image. The next image shows the difference between the s and g distances for the bolted connection.
The area net, while selecting a vertical line AB= Ag – the area of bolts.
1-The bolt diameter as given =3/4 inch, we will refer to Prof. McCormac’s remark, while considering the diameter of the bolts, by adding 1/8 inch to the given diameter of the bolt.
While drilling for the bolts, damage to the material will cause an extra dia of 1/16 inch, plus the 1/16 inch, so the total added will be 1/8 inch.
2-After drawing sections A and B, the width is 8 inches, the thickness is 3/8 inches, and there are two bolts.
We estimate the final diameter of the bolt as (3/4+1/8)=7/8″ inch. The number of bolts is 2.
3-The A gross=3/88=3.00 inch2, for the net area Anet we deduct the area of the 2 holes, then Anet=Ag-sum(d)t=3.00-2(3/8)(7/8)=2.34 inch2. for U=1, then A eff=1*2.34=2.34 inch2.
This is a list of the equations used to estimate the limit state of Yielding and limit state of fracture for tension members.
Problem 3-2-determine the critical net area for a plate.
This is the second problem 3-2, it is required to estimate the critical net area of a 1/2 inch thick plate, for bolts dia =3/4 inch.
We have two paths to failure. The first path is ABCD. In this path, we are going to deduct two holes, and there is no zigzag line.
1-The area gross Ag=111/2=5.50 inch2, for Anet=Ag-sum(d)t=5.50-2(1/2)(3/4+1/8)
=5.5-(7/8)=4.63 inch2.
For the second path, which is a staggered line path ABEF, in this path, we are going to deduct two holes and also consider the zigzag line BE for which S=3″ and g=6″.
2a-The area gross Ag=11*1/2=5.50 inch2.
2b- For the net area for the staggered line ABEF– Anet=Ag-sum(d)t+tplS^2/4g=5.50-2(1/2)(3/4+1/8)+1/2(3^2)/(46)=5.5-1/2*(7/8)+0.1875=4.81 inch2.
For the third path, which is path ABCEF, in this path, we will deduct three holes and consider the zigzag line CF for which S=3″ and g=3″.
3a-the area gross Ag=11*1/2=5.50 inch2.
3b- for Anet=Ag-sum(d)*t+tpl*S^2/4g=5.50-3*(1/2)*(3/4+1/8)+1/2*(3^2)/(4*3)=5.5-1.3125+0.375=4.56 inch2. Finally, the least net area will be equal to 4.56 inch2; the final route selected is ABEF.
That is the end of the problem- 3-2. Thank you all.
This is the pdf used for the illustration of this post.
For more problems, please refer to the following external link.
For the next post, A Solved problem 3-1 for the nominal strength.
For a useful external link-Chapter 3 – Tension Members– Bartlett Quimby.