Feature image for post 17-tension

17-Problem 3-4-3 for MC section-LRFD and ASD value-2/2.

Problem 3-4-3 for MC section-LRFD and ASD value-2/2.

We are dealing with the second part of problem 3-4-3 for MC section-LRFD and ASD value-2/2. We will check the fourth failure line. Decide the least value of the net area. We will estimate the effective area by multiplying it by the factor U. Factor U is obtained from case 2.

We can estimate the nominal strength due to yielding and rupture. finally, we will get the LRFD and ASD values for strength.

Problem 3-4-3 for MC section-4rth failure line-ABFGD’.

We check the fourth failure line which is line ABFGD’, please refer to the next image for more details. Each bolt transfers the applied force by a ratio of force value by the bolt’s number which is equal to P/7. Again we will estimate the sum of deductions due to holes and additions due to staggered bolts. the net value will be multiplied by the web thickness.

We will deduct three holes each hole equals 7/8 inches, and the total deductions equal 21/8 inches.
For the addition we have one zigzag line which is line BF, S value equals 2.5 inches and the g value is the distance between the first and second line of bolts and equals 2 1/2 inches. The net value is (-21/8+0.625)=-2 inches.

The fourth route ABfgd' data.
Problem 3-4-3 for MC section-staggered bolts-ABFGD’.

The net area of the section can be estimated by summing the gross area of the MC section and the area of the net value after multiplying it by web thickness. we can see that the net area will be equal to 6.22 inch2. But this is not the final value.

The section AEFGD’ excludes one bolt that transfers P/7 to the Gusset plate. The section is fit for only 6/7P which is the remaining value of force due to the sum of forces. the modified area will be (6.22*7/6)=7.26 inch^2.

The value of the net area for the fourth route.
The net area of the fourth line of failure.

Problem 3-4-3 for MC section-LRFD and ASD value-2/2- selection of the final net area.

We will list all the values of net areas due to the different routes or the failure lines, as we can see that the least value of the net area is 6.32 inch^2 due to route ABCD. This will be selected to estimate the effective area that has a 100% P load-carrying capacity.

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Problem 3-4-3 for MC section selection of the final net area.

We will estimate the U value from table D3.1 in which the U =1-(x bar/L). X bar is the distance from the CG of the section to the connection, while L is the length of the connection. L value equals 5.5 inches, while fr x bar we need to check the value from table 1-6 for Mc sections.

How to estimate The U value?
How do we estimate the reduction factor U?

Estimate the effective area for the MC section.

We can get the X bar value from table1-6 as we can see from the next image, we have section MC 9×23.9

X bar value equals 0.981 inches for problem 3-4-3 for MC section-LRFD and ASD value-2/2.

We multiply U by the net area to get the effective area value. U=(1-(0.981/5.50)=0.8218. The effective area will be equal to 5.193 inch^2.

Estimate the nominal load due to yielding and rupture.

We can estimate the nominal strength due to yielding and rupture. For yielding the nominal load will be equal to the product of gross area by the yield stress.While for rupture the nominal load will be equal to the product of the effective area by the effective area value. The values of the nominal strength for the two cases are shown in the next slide image.

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Problem 3-4-3 for Mc section-tension members

How to find strength for the MC section?-LRFD and ASD values.

We will estimate the LRFD strength due to yielding and rupture considering the phi value=0.90 for yielding and 0.75 for rupture. We will multiply Pn due to yielding by 0.90. We will multiply the nominal strength due to rupture by 0.75.

The two values are 315.9 kips and 253 kips.

The available design strength is the lower value of the two values due to yielding and rupture, which is 253 kips.

we will estimate the ASD strength due to yielding and rupture considering the omega value=1.67 for yielding and 0.75 2.00 for rupture. We will multiply Pn due to yielding by 0.9(1/1.67). We will multiply the nominal strength due to rupture by (1/2).

This is the final part of Problem 3-4-3 for MC section-LRFD and ASD value, we can get the LRFD and ASD values based on the conditions of yielding and rupture.

The available allowable strength is the lower value of the two values due to yielding and rupture, which are 253,168.80 kips. This is the end of for MC section-staggered bolts.

Problem 3-4-3 for MC section-LRFD and ASD value
Problem 3-4-3 for MC section-LRFD and ASD value

Thanks a lot and see you in another post.

Please refer to the first part of this post, Problem 3-4-3 for staggered bolts-tension members-1/2.

The next post will include, the Effective Area for a Built-Up Section.

For a more detailed illustration of block shear, there is a very useful external link-failure path examples. A Beginner’s Guide to the Steel Construction Manual, 14th ed.

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