## Solved Problems For The Net Area Estimation.

### Brief Content of the video.

We start reviewing the example 3-4-2 from Prof. Salmon’s book and determine the net area of the shown angle where 15/16 inch dia holes are used.

In the unequal Angle 6x4x1/2 inch, the lower g distance is 2 1/2″ as a gauge line and as for the long side of the angle, the first gauge distance is 2 1/4 ” from the edge, as g1 distance, and the second gage distance is 2 1/2 inch between the centerlines of bolts, as g2. The angle dimension in mm, was given between brackets as ( L152 x 102×12.7 mm).

Our way of estimation is as follows, for the zigzag line route, for Anet =Agross- the sum of diameter bolts holes, encountered in your selected route *t the thickness of the element plus the S^2/4g, where S is the horizontal distance in the direction of the applied load, g is the distance perpendicular to gauge lines.

This is a part of the video, which has a subtitle and closed caption in English.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

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

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

### 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, is required to estimate the net area for an unequal angle 6×41/2″, where 15/16″ holes are used.

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″

In order to estimate the net area, we make the following steps:

We have to get the gross area from table 1-7 for our angle 6x4x 1/2″ which is Ag=4.75 inch2.

The edge distance g for AB=2.50″, while g distance for BC =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 is to be ** unfolded for** the evaluation of -g distances based on the centerline distances to the edge as shown in the picture.

For path ac, 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 ABB’, we have a zig-zag line Ab, we have to add( t*S^2/4g), 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.

C-check path ABEE’, we have 3 holes of D=1″, t=1/2″, we have two zigzag lines, but each one has different g distances, but with 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*4.50)= 4.75-1.5+0.70=3.95 inch2.

D-check path ABCC’, we have 3 holes of D=1″, t=1/2″, we have two zigzag lines, but each one has different g distances.

But with the same S distance, so we have to 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 least path is path ac for which A_{net}=3.75 inch2 for the solved problem 3- 4-2.

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

In the Solved problem 3-5, it is required to estimate the net area for a channel C15x 33.90, which has 4 bolts of dia 3/4 inch.

1-Again the C channel is to be unfolded, the g distances are calculated by adding the centerline distance to the edges- web thickness.

2- Get the area of the channel from the relevant table 1-5, Agross=10.00 inch2, as shown in the next picture.

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″. *We have a second zigzag* line CD with s=3″ and g=9″.

the third zigzag line DE with s=3″ and g=4.60″.3- four holes to be deducted with D=7/8″. Two holes at the flange with flange thickness=0.65″.

While 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, DE for the solved problem 3-5, as follows:

For BC & DE, since they’re identical, the thickness is taken as the average of (0.65)+(0.40), since the two lines pass through the flange and the web of the channel, then the tav=1.05/2=0.525″. While for the CD it passes entirely through the web, for which the thickness to be taken as t=0.40″.

The calculation for all the lines is shown as per the next two slide images.

the Anet can then be obtained for the solved problem 3-5, the value is 8.774 inch2.

The final net area data is shown in the next slide image.

This is the pdf data used in the illustration of this post.

This is the next 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.