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
Dr.
Ali I. Tayeh
First Semester
Steel DesignDr.Ali I. Tayeh
Chapter 5 -B
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.
From elementary mechanics of materials, the shearing stress is
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)
This relationship can be illustrated as :
This relationship can be shown in :
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.
Block shear:
below
Block shear:
Block shear:
below
Answer:
Beams
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.
Beams
Beams
Beams
Beams
Quiz
End