Brief data for post 22-steel beam

22-Step-by-step guide to the solved problem 10-2 for beam.

A solved problem 10-2 For Beam Adequancy For Shear.

A solved problem 10-2-video.

In the solved problem 10-2, a given simple beam under the uniformly distributed load of W21x 55. It is required to check the given section W21x 55 adequacy for shear under the given Dead and Live loads. Two designs are estimated are the LRFD and ASD designs are used. In this post, we will use the manual calculation to get the LRFD and ASD values for nominal shear capacities. the video has a closed caption in English, this is a link to Utube.

A solved problem 10-2, check shear adequacy.

This is a solved problem 10-2, from Prof. McCormack’s book, there is a beam, with W section W21x55, Fy =50ksi. It is required to check the adequacy of this beam to shear.

A solved problem 10-2 to check adequacy for beam in shear.

The beam span is 20′ and is subjected to dead loads Wd=2kips/ft, including self-weight and a uniform live load LL of 4 kips/ft.
From table 1-1, we get all the relevant data for W21x55. A section of W21x55 is drawn, as shown in the small sketch on the left side. We can find the value of h and tw. The enclosed depth h , or the distance between flanges less the fillets is equal to d-2kdes will be found as equal to 18.76” while the web thickness tw=0.375″.

The controlling factor for h/tw as given by the AISC  is given by the equation is h/tw=2.24*sqrt(E/Fy), consider E=29000 ksi, Fy=50 ksi, this equation can be re-written as 381.46*1/sqrt(50)=53.90.

We will check the given section value of h/tw; it will be estimated as (18.76/0.375)=50.03, the actual h/tw is less than the controlling factor which is =53.90. Hence the section is adequate for shear.

The next step is to get the nominal shear for the section where Vn=0.60*Fy*(Aw*cv), cv=1, fy=50.0 ksi, A web=d*tw=20.80*0.375=7.80 inch2. The value of Vn=0.60*50*7.80=234.0 kips.

Vult =1.20Vd+1.60Vl, Since Wd is 2 k/ ft, the span is 20 ft, then V ULT d=1.20210=24 kips.
While Vul-ll=1.60(410)=64.0 kips.Thus V ult=24.0+64.0=88.0 kips.

The nominal shear strength for W section for problem 2-22.

For the LRFD calculations for section adequacy for shear stress, we have Cv =1, then φv *Vn=1*234.0 =234.0 kips, while Vul=88.0 kips, the section is safe.

For the ASD calculations for section adequacy for shear stress, we have Ωv =1.5, then (1/Ω)Vn=(1/1.50)234.0 = 156.0kips, while Vt=(210+410)=60.00 kips, then the section is safe.

The factored nominal shear strength.

We will explain the AISC code provision for shear. this part is not included in the video.

AISC specification provision for the design of Members for shear.

Chapter G is the designated chapter for the design of members for shear.

Chapter G of Aisc provision design of members for shear.

Clause G-1 is for the general provisions. G-2 is for members of stiffened and unstiffened webs.

Code provision G1 and G2 for members with stiffened and unstiffened webs.

Cv value determination for shear of steel beams.

 value for I beams.

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

This is a useful external resource for Shear Strength Limit State.
This is the next post, the solved problem-10-2, how to use table 32 for shear value? 

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