## Solved Problem For design strength of tension member.

*Brief content of the video.*

We will start reviewing the solved problem 3-6 for the design strength. For LRFD and ASD design tensile strength, from prof.McCormac handbook.

First, let us review the relevant equations, we have tensile strength by yielding, for which we have Pn=fy*Ag.

For the design according to the LRFD, we will * φt, where φt =0.90. If we wish to evaluate by the ASD, then 1/ Ωt, for which the value is 1/1.67 in tension. For tensile rupture, we have Pn=fult*Aeff, and in the case of LRFD then φt= 0.75. In the case of ASD, then Ωt=2.00.

Where the effective area is estimated U from tables or various values based on the shape of the element is based on the relation: A eff= Anet * U, where, *This 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.

### Limit states for tension members.

Limit state of yielding and rupture and the corresponding values for φt= 0.75 and Ωt=2.00 for each case.

T*his table* includes the various values for Fy and Fult, for ASTM A 36, for ASTM A 572, the grade should be specified Fy= 42 ksi till 65 ksi, for ASTM A514 Fy= 100 ksi, Fy= 100 ksi.

### A Solved problem 3-6.

**Our solved problem 3-6**, Determine the LRFD design tensile strength and the ASD allowable design tensile strength for a W10x45 with two lines of 3/4 inch in diameter bolts in each flange, Using ASTM A572 grade 50, Fy=50 ksi and Fu=65 ksi and the Aisc specification.

Each line has 3 bolts for both the upper and lower flange, Our first step is to get the area of the w section, by using the relevant table For W10-45.

The area Ag=13.30 inch2, the diameter of each bolt is 3/4 inch, then we add 1/8 inch, the overall diameter will be 7/8 inch.

Back to the statement of this solved problem, the bolts are assumed to be at least three bolts, there is an item in the table that states At least three bolts and more, the U value can be obtained via the table.

We can proceed to evaluate the LRFD for ASD tensile strength by Yielding as shown in the following picture, for the Solved problem 3-6.

For LRFD, Ag=3.30 inch2, Fy=50 ksi,φt =0.90, the φt *Pn=0.9*3.30*50=599 kips this is the LRFD Nominal strength for design based on yielding.

For ASD, Ag=3.30 inch2, Fult=50 ksi,1/ Ωt =1/1.67, the 1/ Ωt *Pn=(1/1.67)*3.30*50=399 kips this is the ASD Nominal strength for design based on yielding.

This is how we estimate U for W10x45 by converting to WT 5×22.50, from which we can get the y bar value, which is 0.907, and consider it as x bar in the equation of U=(1- x̅ /L). Use the equation U=(1- x̅ /L), and then substitute by x̅ =0.907, we get U=0.8867.

While if we use the tables for 3 fasteners, with bf/d>2/3, we have U=0.90, which is bigger than the previous value of U, hence can be used.

We have 4 bolts, 2 at the top flange and the other 2 at the lower flange, our dia of the bolt is 7/8 inch, and T flange=0.62″. We can estimate Ant=Agross-4*(7/8)*0.62=13.30-4*7/8*0.62 =11.13 inch2, *Fult=65 ksi, as given in the solved problem.*

For Aeff=Anet*U=11.13*0.90=10.02 inch2. Tensile rupture by LRFD can be estimated as φt*A eff*Fult=0.75*10.02*65=488.30 kips.

While for tensile rupture by the ASD, it can be estimated as 1/Ωt*A eff*U*Fult=(1/2)*10.02*65=325.65 kips.

For the LRFD tensile strength, take the minimum value of the LRFD Limit states for yielding and rupture as follows:

φt*Pn=min of (599, 488.30)=488.30 kips.

While for the ASD tensile strength, take the minimum value of the ASD Limit states for yielding and rupture as follows: (Pn/Ωt)=min(399, 325.6)=325.60 kips.

This is the Pdf file used for the illustration of this post.

For the next post, A Solved problem 4-2 for shear Lag factor U.**Chapter 3 – Tension Members**– A Beginner’s Guide to Structural Engineering is a great external resource.