Tag: Solved problems for the net area.

Two solved problems are included, in the first problem it is required to estimate the net area for a plate of 8 inch width and 3/8 inch thickness and has two rows of bolts of 3/4 inch bolts. In the second problem it is required to estimate the critical net area for 11×1/2″ plate that has three gauge lines with staggered bolts.

  • 5-Solved problems for the net area estimation.

    5-Solved problems for the net area estimation.

    Solved Problems For the Net Area Estimation.

    This is an explanation in detail of the content of the subject.

    The next slide reviews the gauge lines, showing the first gauge line and the second gauge line in the direction of the force. The vertical distance between the gauge lines is g, and the distance between the centerlines of bolts is S, which is called the pitch.

    Pitch and gauge lines for fasteners.

    The first solved Problem 3-4-2 for the net area estimate of a given angle.

    We have two solved problems. The first solved problem, 3-4-2, requires estimating the net area for an unequal angle of 6x4x1/2″, where 15/16″ holes are used.

    Two solved problems, the first solved problem 3-4-2-determine the net area for a given angle

    For the leg 6″,  we have two gauge lines spaced horizontally by s-3″, and g distance =2 1/2″, while g edge distance=2 1/4″.To estimate the net area, we take the following steps:   
    We have to get the gross area from Table 1-7 for our angle 6x4x1/2″, Ag=4.75 inch2.

    Table 1-7 for angles to get the gross area.

    The edge distance g for AB is 2.50″, while the g distance for BC is 2.25+2.50-(1/4+1/4)=4.25″ since we are deducting the thickness of the angle.

    A-check the different path lines and deduct the hole diameter *t, D is  1″ &  t=1/2″. The angle will be unfolded to evaluate -g distances based on the centerline distances to the edge, as shown in the picture. 

    For path 1ACE2, we have two holes of D=1″, t=1/2″, then Anet=Agross- sum(d*t), Anet= 4.75-2*(1″)*1/2″=3.75 inch2.
    B—For path E1ABE2, we have a zig-zag line Ab; we have to add (t*S2/4 g), where S = 3″ and g=2 1/2″; then, for the two holes of D=1″, t=1/2″. Anet=Agross- sum(d*t)+S^2/4g, Anet= 4.75-2*(1″)*1/2″+1/2*(3)^2/(4*2.50)=4.20 inch2.

    Estimate the net area for path ac.

    C-check path E1ABEE2: We have three D=1″ and t=1/2″ holes. We have two zigzag lines with different g distances but the same S distance.
    So we have to add (t*s^2/4*g1)+ (t*s^2/4*g2), then Anet=Agross- sum(d*t)+(t*s^2/4*g1)+ (t*s^2/4*g2).
    Anet= 4.75-3*(1″)*1/2″+(1/2*(3^2/(4*2.50)+(1/2*(3^2/(4*4.25)= 4.75-1.5+0.715=3.96 inch2.

    Estimate the net area for path abee’.

    D-check path E1ABCE2: We have three holes of D=1″ and t=1/2″ and two zigzag lines, but each has different g distances.
    But with the same S distance, we must add (t*s^2/4*g1)+ (t*s^2/4*g2). Anet=Agross- sum(d*t)+(t*s^2/4*g1)+ (t*s^2/4*g2), Anet= 4.75-3*(1″)*1/2″+(1/2*(3^2/(4*2.50)+(1/2*(3^2/(4*4.50)= 3.96 inch2. 

    The selected path is path ac for the net area Anet=3.75 inch2, with the most minor area for the solved problem 3- 4-2.

    Estimate the net area for path abee’.Estimate the net area for path abee’.

    A Solved problem 3-5 for the estimate of net area for C channel.

    In Solved Problems 3-5, the net area for channel C15x 33.90, which has four bolts with a diameter of 3/4 inch, must be estimated.
    1-The C channel is to be unfolded. We will deduct 2tw thickness from the sum of its external dimensions.

    A solved problem 3-5 find the net aarea for a given C15x33.9.

    2- Get the area of the channel from the relevant table 1-5,  Agross=10.00 inch2.

    The I shape consists of two thickened portions for the flange, with width=tf-tw=3″ and thickness equal to 0.65″.

    The middle portion is the web with a height of 15 inches and a width equal to 0.40 inches.

    The details of the I shape for the C channel.

    I have drawn the elevation of the I shape, which has four gauge lines. The vertical distance between the second and third gauge is 9″. The Other gauge distances are calculated by subtracting 9 inches from the overall depth and dividing by 2.

    The details for the gauge lines for the I shape.

    A—Examine the route ABCDE; for the solved problem 4-2, we have one zigzag line BC with S = 3″ and g=3+2-1*tweb=5-0.40=4.60″ and a second zigzag line CD with S = 3″ and g=9″.

    The third zigzag line DE has s = 3″ and g=4.60″.3—four holes to be deducted with D=7/8″. There are two holes at the flange, with flange thickness=0.65″.

    The other two holes are located at the web, where the web thickness=0.40″.
    We can estimate the whole area to be deducted as 2*(7/8)*(0.65+0.40)=1.84 inch2.

    We can estimate the ( t*S^2/4g) terms for lines BC, CD, and DE for the solved problem 3-5, as follows:
      Since BC and DE are identical, the thickness is taken as the average of (0.65)+(0.40). Since the two lines pass through the flange and the channel’s web, the tav=1.05/2=0.525″.

    While the CD route passes entirely through the web, for which the thickness is to be taken as t=0.40″. The final net area is equal to 8.78 inch2.

    The final calculations for the net area.

    Practice Quiz.

    Quiz 3-18

    This is the following post: The Definition of the Effective Area for Tension Members.
    Chapter 3 – Tension Members– A Beginner’s Guide to Structural Engineering is a great external resource.

Solved problems for net area estimation. 

How do we estimate the net area for plates?

A solved Problem 3-1.

In solved 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, 8 inches by 3/8 inches, connected with two thin plates. Each has an exact width of 8 inches and a 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 seen in the next slide image. The next image shows the difference between the s and g distances for the bolted connection.

Pitch and gauge lines for fasteners.


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. There are two bolts.

3-The A gross=3/88=3.00 inch2, for the net area Anet we deduct the area of the two 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.

Solved problem 3-1 for net area estimation for a given plate.

This is a list of the equations used to estimate the limit state of Yielding and limit state of fracture for tension members.

Tensile rupture and Tensile yielding  Tensile rupture and Tensile yielding equations.-post 2equations.

Problem 3-2-determine the critical net area for a plate.

This is the second problem 3-2. It requires estimating the critical net area of a 1/2-inch thick plate for bolts with a diameter of 3/4 inch.

We have two paths to failure. The first path is ABCD. In this path, we will 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.


Solved problem 3-2 for determining the critical net area for a given plate.

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.
 

Net area for the first route ABEF.

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 equal 4.56 inch2; the final route selected is ABEF.
 

The calculation for the least net area.

That is the end of the problem- 3-2. Thank you all.
For more problems, please refer to the following external link.
For the next post, A Solved problem 3-1 for the nominal strength.

See Chapter 3, Tension Members, and Bartlett Quimby for a useful external link.