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
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter6: Beams And Cables
Section: Chapter Questions
Problem 6.42P: For the beam AB shown in Cases 1 and 2, derive and plot expressions for the shear force and bending...
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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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