Category: Deflection

Introduction to deflection of steel beams: How can we estimate the required moment of inertia based on deflection?

  • List of steel Beam Posts-part-3b.

    List of steel Beam Posts-part-3b.

    List of Steel Beam Posts-part-3b.

    Deflection of steel beams.

      

    The post is part of a list of steel Beam posts-part-3b, including deflection disadvantages.
    2-The permitted limits of deflection.
    3- What are the Deflection limitations in the unified building code?
    4- What is the C1 coefficient as a deflection parameter by the AISC?

    The required deflection values as a ratio of the span length.

    C1 value used by the CM#15 for deflection equation.

    Here is the link for post 24: Deflection of steel beams.

    A solved problem 5.1 from Prof. Fredrick Roland’s book. Example 5.1-part 1.

    The 30-foot beam is laterally supported for its entire length. It supports a uniform dead load, including a beam weight of 0.30 kip/ft and a uniform live load of 0.70 kip/ft. In the first part, we will find the required moment of inertia value Ix based on the Total load and Ix based on the LL.

    Solved problem 5.1-estimate the ultimate moment.

    Find the requied Ix.

    Here is a link to post 24a– Easy approach to the solved problem-5-1-part 1-2.

          

    A solved problem 5.1 from Prof. Fredrick Roland’s book- part 2. 

    In part 2, we will estimate the ultimate moment in the LRFD design. We will use Table 3-2 to select a W section and check the Ix for the chosen section against Ix values based on deflection. We will use Table 3-3 to select the proper section based on Ix. Check the section chosen for The ASD design.

    Use table 3-3 for Ux to select a W section.

      Here is a link to post 24b– Easy approach to solved problem-5-1-part 2-2.

    Part 1/4 Solved problem 9-9-6: how to find LL?

    Part 1 includes a solved problem,9-9-6, from Prof. Charles G Salmon’s book. Determine the service live load the beam may carry if the Dead load is  0.15 kip/ft, including the beam weight. Use load and factor design. The steel has Fy=65ksi. The given beam is a built-up beam. We will use Table b4.1B to determine the limiting value for flange and web parameters for local buckling.

    The first part of the solution includes determining the lambda coefficients λp-F λr-F, λp-w λr-w for Local buckling for the Flange and the web of the Built-up section.

    Both the flange and web sections are Non-Compact sections since the limiting parameters are less than those of lambda Plastic.

    estimate of lambda plastic for flange,lambda r for flange.

    Here is the link to post 25 –How to find LL- Part 1/4?

    Part 2/4 for the Solved problem 9-9-6, how to find LL?

    The second part of the solution includes determining the section modulus of elasticity for the Built-up section. Two nominal moments: one is based on the limit state of local buckling for Flange, and the other is based on the local buckling of the web section. After checking the lateral torsional buckling, the final decision on the final value of the nominal moment will be made, which will be included in the next two parts.

    The nominal moment for the Built-up section.

    Here is the link to post 26- Part-2-4 of the Solved problem 9-9-6, How To Find LL?

    Part 3/4 for the Solved problem 9-9-6, how to find LL?

    The third part of the solution includes the lateral-torsional buckling parameters Lp and Lr for the built-up section, estimating the value of limiting length lr.

    The value of Lr.

    Here is the kink to post 27: Part 3/4 of the Solved problem 9-9-6: How do I find LL?

    Part 4/4 for the Solved problem 9-9-6, how do we find LL?

    The fourth part of the solution includes evaluating the nominal moment for the section based on the Lateral torsional buckling and then finalizing the LRFD value of the Final nominal moment to estimate the permitted Live load value.

    The final live load  for the Built-up section.

    Here is the link to post 28:  Part 4/4 Solved problem 9-9-6. How do you find LL?

    The next post will be the list of steel beam posts, part 4.

    A very useful external resource, A Beginner’s Guide to Structural Engineering



  • 24b- Easy approach to solved problem-5-1-part 2-2.

    24b- Easy approach to solved problem-5-1-part 2-2.

    Solved problem-5-1-part 2-2.

    Solved problem-5-1-part 2-2. is the title of this post, in which we continue solving problem 5.1; we have obtained the required value of the moment of inertia, Ix, as equal to 440 inch4. We have selected W14x30 based on a Zx value of 44.40 inch3; we need to check its Ix value from Table 1-1. The Ix value equals 291.0 inch4, which is smaller than 440.0 inch4. So, we need to revise the selection.

    Solved problem-5-1-part 2-Details of W section W14x30

    Use Table 3-3 to select a new W section with Ix>or equal to 440 inch4.

    Select new W section.

    Use table 3-3 for solved problem-5-1-LRFD design.

    Use Table 3-3 to select a new W and use the bold W18x35. This gives Ix equal to 510 inch4, greater than 440 inch4. We use Table 1-1 to get the value of Zx for w18x35. The Zx value is equal to 66.50 inch3.

    We will estimate the design strength for W18x35 based on the LRFD design. It will equal 249 Ft. Kips, which is bigger than the Ultimate moment, which equals 166.50 ft. kips. The section W18x35 is adequate for design.

    Table 3-3 to select W18x35.

    Find the required Zx based on the ASD design.

    We need to estimate the required Zx for the beam based on the ASD design. The total moment equals 112.50 ft, kips, and the required Zx value equals 45.09 inch3

    Estimate the required Zx for beam based on ASd design.

    Due to the inertia requirement, the Ix value is 440 inches4, and the required Zx is 45.09 inches3; we need to use Table 3-3 based on Ix to find the proper section.

    The need of use of table 3-3 for ASD design.

    Use Table 3-3 for solved problem-5-1-ASD design.

    Use Table 3-3 to select a new W and use the bold W18x35. This gives Ix equal to 510 inch4, which is greater than 440 inch4.

    We have a total moment equal to 112.50 kips. Check the allowable strength for W18x35, which is equal to (1/omega)*Mt/Zx*Fy, which is equal to 166.0 Ft. kips. This value is bigger than the Total moment. The section is adequate for design based on the ASD design. Thanks a lot.

    The final selected section based on table 3-3-ASD

    As an external resource for the deflection Chapter 8 – Bending Members
    For the next post, 25-part-1-4-solved-problem-9-9-6-1-4, how to find LL for a given slender section?  

  • 24a- Easy approach to the solved problem-5-1-part 1-2.

    24a- Easy approach to the solved problem-5-1-part 1-2.

    Solved problem-5-1-part 1-2.

    Solved problem-5-1-part 1-2 for deflection is the title of this post, which is a design example for a simply supported beam.

    In this post, Deflection of Steel Beams Part 2, we will select the beam’s W section based on the Zx value.

    A solved problem-5-1-part 1-2 for the design of a beam.

    The solved problem 5-1 is quoted from the Book of Prof. Fredrick Roland’s book.
    Zone -1 bending has M plastic moments, for which Mp=Fy*Zx; the plastic theory is discussed in previous posts.

    We have a supported beam with a span of L=30 ft. The steel is ASTM A992, where Fy=50 Ksi and Fult=65 ksi. The dead load is 0.30 kip/ft, and the live load is 0.70 kip/ft.
    This solved problem is a design example with given deflection criteria.

    This problem was selected after reviewing the previous information for the LRFD and the ASD for the design under bending.
    The beam is laterally supported for its entire length. Therefore, the section will be considered compact and in zone 1.

    If we start with the LRFD design, we need to estimate the beam’s moment as M=W*L^2/8, cb=1, which is the moment coefficient. The section is fully braced.

    Factored Load for LRFD design and the ultimate moment.

    First, the Ultimate load =1.20D+1.60 L for D=0.30 kips/ft and L=0.70 Kips/ft, then Wult=1.20*0.30+1.60*0.70=1.48 Kips/ft, The value of Mult will be equal to 1.48*30^2/8=166.50 ft. kips. To convert to Inch kips, multiply by 12, 66.50*12= 1998 inch—kips in the LRFD design.

    Solved problem-5-1-part 1-2.

    How do we get Zx, which is based on LRFD?

    The reduction factor for strength based on the lRFd Φb=0.90 while Ωb=1.67, to check Φb* Ωb=1.50. For step-1. We will estimate Zx from the known formula Mn=Fy*Zx, Fy=50 ksi, then Φb*Mn =Φb*Fy*Zx=Mult, The yield street Fy=50 ksi. The plastic section modulus Zx= Mult/0.90*50* Zx=1998/45=44.50 inch3.

    Detailed estimate for Zx based on LRFD design.

    Once we have estimated the required Zx, we will get the necessary W section based on Table 3-2, which arranges sections based on Zx. The first selection is W14x30, but we must check the inertia based on serviceability requirements.

    use table 3-2 to find the proper W section based on Zx.

    Determine the moments based on Live load and Moment for total load.

    We will estimate the total load, the sum of D plus L, to equal 1.0 kip/Ft. The beam’s momentum based on the total load equals 1*30^2/8=112.50 Ft.kips; please refer to the next slide image.

    The Moment value based on total load acting on the beam.

    We estimate the live load moment to be 0.7 k/ft, which equals 78.75 ft. kips.

    We will need both moment values to find the deflection values.

    Inertia for the beam to give a safe deflection case of D+L.

    Deflection is based on Dead and live loads and will be estimated based on where the deflection delta equals L/240. The length is 30 feet. We convert to inches by multiplying by 12.

    The deflection value due to the total load Δ-L=L/240=360/240=1.5″. The moment value equals 112.50 Ft.kips, which we estimated earlier.

    Use the AISC relation for deflection as equal to ML^2/C1*Ix, where c1=161, equate M*L^2/C1*Ix to 1.50 inch. M is in Ft. Kips, L in Feet. We will get the Ix value as equal to 419.25 inch4.

    The required moment of inertia based on deflection due to total load.

    Inertia for the beam to give a safe deflection case of L.

    Checking the deflection of the steel beam based on L/360 for L is required.
    We are in the serviceability criteria; we use L=0.70 kips/ft. The deflection due to total live load equals (30* 12/360)=1.00 inch.

    The moment used is the value due to the live load equal to 78.75 Ft.kips. Use the AISC relation for deflection as equal to ML^2/C1*Ix. We will get the Ix value as equal to 440.22 inch4. Please refer to the next slide image for more details.

    The moment of inertia that is required based on the deflection due to live load.

    The final selected Inertia for the beam.

    We have two values: the first one is 440.22 inch4, and the second estimated inertia value is 419.25 inch4; we will select the bigger value, which is 440 inch4.
    .

    The final selected moment of inertia.

    In the next post, we will continue our estimation for selecting the final economic section.

    As an external resource for the deflection. Chapter 8 – Bending Members.
    This is the next post, A solved problem 5-1 part 2-2, how to find LL for a given slender section?  

  • 24- Easy approach to deflection of steel beams.

    24- Easy approach to deflection of steel beams.

    Deflection of steel beams part 1.

    This is the content of the lecture. What are the deflection parameters?

    Content of post 24.

    What are the disadvantages of the deflection of steel beams?

    This is a quote from Prof. Mccormac’s book, chapter 10. One of the disadvantages of deflection is that it may damage other materials and create plaster cracks. Therefore, the deflection is usually limited to a maximum value since it

    The deflection may damage other materials attached to or supported by the beam in question. The second point is that structures are often damaged by excessive deflections, which give them a bad appearance.

    The third point is loss of confidence due to the appearance of deflection for a building. The last point is that it may be necessary for several different beams supporting the same loads to deflect equal amounts for appearance purposes to be the same amount of deflection.

    Disadvantages of excessive deflection.

    What are the parameters of the deflection of steel beams?

    The live load deflections are approximately equal to L/360 of the span length. However, the reader should note that the limitation of the deflection of steel beams falls in the serviceability area.

    Therefore, deflections are determined for the service loads, so we cannot use the Ultimate loads, such as the 1.2D+1.60L combination of loads, to estimate the deflection.

     What tools enable us to determine deflection? We use the moment area method, converting the area of moments into loads from which we estimate the shear forces, which gives us the value of the slope.

    From the bending produced from Moment loads, we can evaluate the deflection values and use the conjugate beam method, which converts the end cantilever portion into fixed support and the hinged end into hinged supports.

    he deflection values are based on the service loads.

    The third method is virtual work. The deflection value = 5WL^4/384*EI for a simply supported beam under uniform load.

    Limits of deflection values.

    IBC provision for deflection.

    The IBC, the building international building code, in which the beam deflection values are grouped into three categories, was Quoted from Prof. Fredrick Roland’s book Steel Design for Civil P.E.

    The first item is for the roof members: Live load L& snow load S or wind load W, and the third category is the dead load plus live load D+L for each category, roof members if supporting plaster ceilings, and the deflection value.
    Deflection is permitted for live load L to equal L/360, where L is the span. For snow Load s or wind load W, the deflection equals L/360; for D+L, the deflection is L/240.

     For Roof members supporting non-plaster ceilings, the deflection values increased and became L=L/240 for live load, while for snow Load or wind load W=L/240 for D+L, the deflection is L/180.

    For floor members, we do not have snow loads.

    The allowable deflection is permissible under Live Loads L =span/360, while for D+L loads, the deflection will be considered as ( span/240).

    Deflection limitation values based on the IBC.

    This is the provision of the Aisc specification for Deflection based on serviceability.

    What is factor C1 used in the deflection equation by AISC?

    In the new section in the AISC, a new coefficient was given as a C1 factor to account for the errors due to the different factors; the deflection is delta =M*L^2/(C1*Ix), so instead of using a deflection value of (5/384)w*L4/EI, we consider instead, M as (W*L^2/8), L as span length in inches, w in kips/ft and E as 29000ksi, we can get the c1 value.

    The next image explains how we get the value of c1 for the deflection of a supported beam under uniform loading. Covert the term ML^2 units to kips inch^3 by multiplying the values by 12^3. C1 for the case of a simple beam under a uniform load will be equal to 48*29000/12^3*5=161.11.

    Using parameters for the deflection values of steel beams.

    We will continue the subject in the next two posts. A solved problem will be presented.

    As an external resource for the deflection. Chapter 8 – Bending Members.
    This link for the next post solved problem 5-1-part 1-2.