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5-Solved problems for the net area estimation.

Solved Problems For The Net Area Estimation.

Brief Content of the video.

We start reviewing 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 millimeters 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.

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 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, is required to estimate the net area for an unequal angle of 6×41/2″, 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″.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.

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

The edge distance g for AB=2.50″, while the 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.

Estimate the net area for path ac.

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.

Estimate the net area for path abee’.

D-check path ABCC’, we have 3 holes of D=1″, t=1/2″, and 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. 

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

The selected path is path ac for which Anet=3.75 inch2, with the least area 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-The C channel is to be unfolded, We will deduct 2tw thickness for the sum of the external dimensions of the C channel.

Solved problem 3-5-determine the net area for a given channel.

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 the thickness is 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.
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 distance between the second and the third gauge is given as 9″.The Other gauge distances are to be calculated by subtracting 9 inches from the overall depth and dividing by Two.

The details for the gauge lines for the I shape.
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″.   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, and 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 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.
The final calculations for the net area.

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.

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