load is applied to a beam as shown. Determine the equivalent force and location of the distributed load when a = 6 ft, b = ft, WA = 8 kips/ft and wg = 13 kips/ft WA A A a Parabola Vertex B WB
load is applied to a beam as shown. Determine the equivalent force and location of the distributed load when a = 6 ft, b = ft, WA = 8 kips/ft and wg = 13 kips/ft WA A A a Parabola Vertex B WB
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:A distributed load is applied to a beam as shown. Determine the equivalent force and location of the distributed load when a = 6 ft, b = 7 ft, ωA = 8 kips/ft and ωB = 13 kips/ft.
**Diagram Explanation:**
- The diagram depicts a beam AB with a distributed load varying linearly across its length.
- The distributed load starts at point A with intensity \( \omega_A = 8 \) kips/ft and increases in a parabolic manner to \( \omega_B = 13 \) kips/ft at point B.
- The length from A to a certain point is marked as 'a' and from this point to B is denoted as 'b'. In this case, \( a = 6 \) ft and \( b = 7 \) ft.
- The load distribution forms a parabolic curve with its vertex along the beam.
**Calculation Requirements:**
The equivalent force is calculated as the area under the load distribution curve. Its location is determined using the centroid of the load shape.
_Fill in the blanks:_
The equivalent force is __ kips (Click to select) located __ ft left of point B.
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