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LRFD - Steel Design. Dr. Ali I. Tayeh First Semester. 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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LRFD-Steel Design

Dr.

Ali I. Tayeh

First Semester

Steel DesignDr.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

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.

Beams-shear

From elementary mechanics of materials, the shearing stress is

Beams-shear

the relationship between shear strength and the width-thickness ratio ia analogous to that between flexural strength and the width thickness ratio (for FLB or WLB) and between flextrual strength and unbraced length (for LTB)

Beams-shear

This relationship can be illustrated as :

Beams-shear

This relationship can be shown in :

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.

Beams-shear block

Block shear:

• "Block Shear Rupture Strength," gives two equations for the block shear design strength:

Beams-shear block

Block shear:

• Example 5-8

Beams-shear block

Block shear:

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

Beams- Deflection

• Deflection limit:

Beams- Deflection

• Example 5-9 :

• solution

Beams- Deflection

• solution

• 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.

• DESIGN:

• Example 5.10:

Answer:

• DESIGN:

• Example 5.10 cont:

• DESIGN:

• Example 5.10 cont:

• DESIGN:

• Example 5.10 cont:

• 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.

• DESIGN:

• DESIGN:

• DESIGN:

• 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.

• DESIGN:

• Example 5.11 :

• DESIGN:

• Example 5.11 :

• DESIGN:

• Example 5.11 :

• 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.

• DESIGN:

• Example 5.12.cont

• DESIGN:

• Example 5.12.cont

• DESIGN:

• Example 5.12.cont

• DESIGN:

• Example 5.12.cont

• DESIGN:

• Example 5.12.cont

• 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.

• 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.

• Bearing plates and column base plates :

• Bearing plates and column base plates :

• Web Yielding

• At the interior load, the length of section subjected to yielding is:

• Bearing plates and column base plates :

• Web Yielding

• The nominal strength is:

• The design strength is ØRw where Ø = 1.0

• Bearing plates and column base plates :

• Bearing plates and column base plates :

Beams -Steel Design

End