) A 20 ft horizontal beam is attached to a wall with a fixed support. If the beam is subjected to the distributed loads indicate
1 answer:
Answer:
Ra = 3,000 lb
Ma = 26,125 lb-ft
Explanation:
Solution:-
- We are given a cantilever horizontal beam with a span of 20 ft subjected to distributed loads.
- We will use the following conventions and define our coordinate system:
- Denote positive direction ( + ) directed "vertically upward" and "anticlockwise".
- Denote negative direction ( - ) directed "vertically downward" and "clockwise".
- We will denote a reaction force ( Ra ) and reaction moment ( Ma ) exerted by the wall support at point A on the beam in positive directions, respectively. These reactions are consequence of the distributed load acting on the horizontal beam.
- First step is to analyze the distributed load. We will make 4 sections of the distributed loads and write down the shapes of each distributions as follows:
Section Load ( Lb/ft) Distribution
1 750 Rectangular
2 750 Right angle triangle
3 450 Right angle triangle
4 450 Rectangular
- We will proceed with load-transformation. In load transformation we convert the distributed load into a " point load " and determine the "location of transformed load " using distribution pattern of each section.
- We will now determine the point load and location of each point load for the respective section from point ( A ) as follows:
Section 1:
P1 = Load_1 * distribution span_1
From above table, Load_1 = 750 lb/ft.
Distribution is rectangular, spanning L1 = 3ft
P1 = ( 750 lb/ft ) * ( 3ft )
P1 = 4500 lb
For rectangular distributions, the point load acts at the geometric center of the distribution pattern; hence, center of rectangle x1 = 3ft from point A.
Section 2:
P2 = Load_2 * distribution span_2*0.5
From above table, Load_2 = 750 lb/ft.
Distribution is right angle triangle, spanning ( L2 )
P2 = 0.5* ( 750 lb/ft ) * ( L2 )
P2 = 375*L2 lb
For rectangular distributions, the point load acts at the geometric center of the distribution pattern; hence, center of right angle triangle is x2 = 6ft + L2/3 from point A.
Section 3:
P3 = Load_3 * distribution span_2*0.5
From above table, Load_3 = 450 lb/ft.
Distribution is right angle triangle, spanning ( L3 )
P3 = 0.5* ( 450 lb/ft ) * ( L3 )
P3 = 375*L3 lb
For rectangular distributions, the point load acts at the geometric center of the distribution pattern; hence, center of right angle triangle is x3 = 6ft + L2 + 2*L3/3 from point A.
- Before we proceed further, we need to determine the spanning lengths of right angle triangles for distributed loads for sections 2 and 3. From the given figure we can form two equations as follows:
L2 + L3 = 9 ft .... Eq 1
- The second equation can be determined by the concept of similar triangles to relate L2 and L3 as follows:
L2 = L3* ( 750 / 450 )
L2 = 5*L3 / 3 ... Eq 2
- Solve the two equations simultaneously for L2 and L3, we get:
L2 = 4.625 ft, L3 = 4.375 ft
- The corresponding point loads and locations are as such:
P2 = 1734.375 lb , x2 = 7.5416667 ft
P3 = 984.375 lb , x3 = 13.541667 ft
Note: The division of 750 lb/ft and 450 ft/lb loads in section 2 and section 3 can be mistaken to be equal and setting L2 = L3 = 4.5 ft. Remember to take care of such situations.
Section 4:
P4 = Load_4 * distribution span_4
From above table, Load_4 = 450 lb/ft.
Distribution is rectangular, spanning L4 = 5ft
P4 = ( 450 lb/ft ) * ( 5ft )
P4 = 2,250 lb
For rectangular distributions, the point load acts at the geometric center of the distribution pattern; hence, center of rectangle x4 = 6ft + 9ft + L4/2 = 17.5 ft from point A.
- Second step: Apply the static equilibrium conditions on the horizontal beam.
Force Balance: Sum of forces in vertical direction = 0
∑
Moment Balance: Sum of moments about A = 0
∑
Answer: The reaction force at point A is directed vertically upward ( Ra = 3,000 Lb ) and the reactive moment is directed clockwise at point A ( Ma = 26,125 Lb-ft ).
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