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LRFD- Steel Design

LRFD- Steel Design. Dr. Ali Tayeh. Second Semester 2010-2011. Steel Design Dr. Ali I. Tayeh. Chapter 5 -B. Beams- shear. Where: Vu = maximum shear based on the controlling combination of factored loads Ø = resistance factor for shear = 0.90 Vn = nominal shear strength.

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LRFD- Steel Design

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  1. LRFD-Steel Design Dr. Ali Tayeh Second Semester 2010-2011

  2. Steel DesignDr.Ali I. Tayeh Chapter 5 -B

  3. Beams-shear Where: Vu = maximum shear based on the controlling combination of factored loads Ø = resistance factor for shear = 0.90 Vn = nominal shear strength • Consider the simple beam as shown. At a distance x from the left end and at the neutral axis of the cross section, the state of stress is as shown. Because this element is located at the neutral axis, it is not subjected to flexural stress.

  4. Beams-shear From elementary mechanics of materials, the shearing stress is

  5. Beams-shear the relationship between shear strength and the width-thickness ratio is analogous to that between flexural strength and the width thickness ratio (for FLB or WLB) and between flexural strength and unbraced length (for LTB)

  6. Beams-shear This relationship can be illustrated as :

  7. Beams-shear This relationship can be shown in :

  8. Beams-shearblock Block shear: To facilitate the connection of beams to other beams so that the top flanges are at the same elevation, a short length of the top flange of one of the beams may be cut away, or coped. If a coped beam is connected with bolts as in Figure below, segment ABC will tend to tear out. along line AB and there will be tension along BC. Thus the block shear strength will be a limiting value of the reaction.

  9. Beams-shear block Block shear: "Block Shear Rupture Strength," gives two equations for the block shear design strength:

  10. Beams-shear block Block shear: Example 5-8 below

  11. Beams-shear block

  12. Beams-shear block Block shear:

  13. Beams- Deflection Deflection: For the common case of a simply supported, uniformly loaded beam such as that in Figure below, the maximum vertical deflection is

  14. Beams- Deflection Deflection limit:

  15. Beams- Deflection Example 5-9 : solution below

  16. Beams- Deflection solution

  17. Beams DESIGN: Beam design entails the selection of a cross-sectional shape that will have enough strength and that will meet serviceability requirements. The design process can be outlined as follows: Compute the factored load moment, Mu. it will be the same as the required design strength, ØMn. Select a shape that satisfies this strength requirement. This can be done in one of two ways: Assume a shape, compute the design strength, and compare it with the factored load moment. Revise if necessary. The trial shape can be easily selected in only a limited number of situations Use beam design chart in part 5 at manual. Check the shear strength. Check the defection.

  18. Beams DESIGN: Example 5.10: Answer:

  19. Beams DESIGN: Example 5.10 cont:

  20. Beams DESIGN: Example 5.10 cont:

  21. Beams DESIGN: Example 5.10 cont:

  22. Beams DESIGN: Beam Design Charts: Many graphs, charts, and tables are available. for the practicing engineer, and these aids can greatly simplify the design process. particularly the curves of design moment versus un braced length given in Part 5 of the Manual. These curves will be described the graph of design moment ¢bMn a function of un braced length Lb for a particular compact shape.

  23. Beams DESIGN:

  24. Beams DESIGN:

  25. Beams DESIGN:

  26. Beams • DESIGN: • Example 5.11 : The beam shown must support two concentrated live loads of 20 kips each at the quarter points. The maximum live load deflection must not exceed L/240. Lateral support is provided at the ends of the beam. UseA572 Grade 50 steel and select a W-shape.

  27. Beams • DESIGN: • Example 5.11 :

  28. Beams • DESIGN: • Example 5.11 :

  29. Beams • DESIGN: • Example 5.11 :

  30. Beams DESIGN: Example 5.12 Use A992 steel and select a rolled shape for the beam shown below. The concentrated load is a service live load, and the uniform load is 30% dead load and 70% live load. Lateral bracing is provided at the ends and at mid span. There is no restriction on deflection.

  31. Beams DESIGN: Example 5.12.cont =

  32. Beams DESIGN: Example 5.12.cont

  33. Beams DESIGN: Example 5.12.cont

  34. Beams DESIGN: Example 5.12.cont

  35. Beams DESIGN: Example 5.12.cont

  36. Beams Bearing plates and column base plates : The function of the plate is to distribute a concentrated load to the supporting material The function of the plate is to distribute a concentrated load to the supporting material. Two types of beam bearing plates are considered One that transmits the beam reaction to a support such as a concrete wall One that transmits a load to the top flange of a beam.

  37. Beams • The design of the bearing plate consists of three steps. • Determine dimension N so that web yielding and web crippling are prevented • Determine Dimension B so that the area N ×B is sufficient to prevent the supporting material from being crushed in bearing. • Determine the thickness t so that the plate has sufficient bending strength.

  38. Beams Bearing plates and column base plates :

  39. Beams Bearing plates and column base plates : Web Yielding At the interior load, the length of section subjected to yielding is:

  40. Beams Bearing plates and column base plates : Web Yielding The nominal strength is: The design strength is ØRw where Ø = 1.0

  41. Beams Bearing plates and column base plates :

  42. Beams Bearing plates and column base plates :

  43. END ANY QUESTION?

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