 # 9-Two Solved problems for an estimate of effective net area.

## Two Solved problems for an estimate of effective net area.

### Brief content of the video.

The first part is an extract from Prof. Segui’s book, for which he gave examples of the various shapes, angles, with two or more fasteners per line, and angles with 4 or more fasteners per line, as well as two W shapes of a different dimension each of the two, has a different bf/d value.

I wrote the dimension of these w sections from the relevant tables, and accordingly the estimated bf/d, and I referred to the table of shapes for the different equations for U factor value, as for the Wt, it is a  section cut from a parent w.

The second part of the video includes the solved example Number# 3.4, for a given angle 6×6 x1/2 inch, bolted with two lines, for each line, there are 3 bolts with 5/8 inch bolt, it is required to estimate the effective area? the video has a subtitle and closed caption in English.

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

### Practice examples for determining the U value for different shapes.

Refer to shapes as per figure 3.10 from prof. Segui handbook. He makes examples of the different shapes, and their u values, for example for single or double angles with two or more fasteners.

We will have different shear lag factors, this is derived from the table for the single angle, or double angles, when the angle is connected to a plate, for which we have two different U values if we have at each line three or more fasteners, then U =0.80.

But if two or three fasteners then U=0.60. Please refer to item #8.

The first figure is for two or three fasteners. The next shape is for four fasteners and more than U=0.80, for W sections, he gave different sections.

For example w10x19  the w section is bolted at both upper and lower flange with four bolts, then u=0.85, if we refer to the table for w sections, and write the breadth of flange bf=4.02 inch and overall depth =10.2 ” and overall depth =10.2 ”.

Estimate the bf/d, If bf/d>2/3, then U=0.90, but if bf/d<2/3, then U=0.85, for the case of  w10x19, we have bf/d is =0.394 <2/3 then U=0.85 this rule is from the table. Please refer to item #7.

For W, M, S or Hp shapes, having bolts at the top and bottom flanges With three or more fasteners, this is the rule if the bf > 2/3*d, 0r bf/d=2/3 then U=0.90, but if the bf< 2/3*d,  then U=0.85.

For the w section if for w 10x 19 the upper flange has 4 bolts at top and bottom, bf/d=0.394< 2/3 then U=0.85, for the other case with w8x24, the depth of the beam is small, the beam is shallow, d is 7.93 ”, if we divide bf/d we get 0.82> 2/3, then U=0.90, this is the case in the table for w section.

If we have Wt , which means it is a T shape, cut from a W section, if we refer to the property for wt 5x 22.5 from the table we have the bf=8.02″ and  d=5″ if we divide bf/d=8.02/5.05, we will find that the ratio will be >1.

What is the parent section? The parent section is W10x45. the depth is twice the Wt section, then bf/d=8.02/10.1=0.794, then the U value =0.90.

In the last case if we have a W shape bolted with two lines of two lines with 4 or more fasteners per single line but these fasteners are at the web, this is the case shown in the table, the U value is =0.70.

The next tables for u value are enclosed for quick reference to match with the examples given by Prof. Segui which is presented by the previous slide image.

Case6a and 6b, for a rectangular Hss, welded to a gusset plate at the middle.

### The first solved problem 3-4 of the two solved problems for an estimate of effective net area.

In the first solved problem-3-4 Determine the effective net area for the tension member in figure 3-12, the given angle is 6×6 x1/2″, I put the corresponding u value for the angle with fasteners.

If we refer to table 1-7 for the angle 6x6x1/2″,  for the solved problem 3-4, we will have y̅ value is 1.67″, which will be considered as x̅ the vertical distance to the connection line or the force line. We need to estimate the net area and U factor, then multiply to get the effective net area.

We have two lines of bolts and each line has 3 bolts, each bolt dia is 5/8″, we add 1/8″ then the d hole=6/8″, for the net area determination, the sum of d *t, where t is the leg thickness, where t is the leg thickness, which 1/2.then Area net=2*(6/8)*(1/2)=6/8 inch2, This is deduction value from the Area gross.

In the next step, The u value to be estimated U=(1-x̅/L), L is the length, for x̅ we have obtained from the table, while the L value is from the first fastener, bolt, to the last one which is=3*2=6″, then U= (1-(1.67/6)) =0.7217.

Our gross area value is   5.77 inch2, we will deduct (6/8) inch2, then Area effective=5.03 inch2 to get the net area, multiply by the A-effective=5.03*0.7217=3.623 inch2.

### The second problem 3-5 of the two solved problems.

Refer to solved problem 3-5, if the tension member of problem 3-4  is welded as shown in fig 3-13, determine the effective area, from the last time, A effective was= 3.623 inch2.

The same equation for U value is used with the same x̅ but this length, average length of welds, for 1/2(5.5+5.5)=5.5 “, U=1-(1.67/5.5)=0.696, but since no deduction due to bolts, the Anet=A gross, then the area effective Aeff=U*Anet=U*A-gross.

The value of the effective area, A eff=5.77*0.696=4.02 inch2. rupture tensile value, in this case, will be more than the case of bolts since A effective is bigger.

This is the PDf file used in the illustration of this post.

The next post is post #10, about the Introduction to block shear.

Chapter 3 – Tension MembersA Beginner’s Guide to Structural Engineering is a great external resource.

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