Question

) A 20 ft horizontal beam is attached to a wall with a fixed support. If the beam is subjected to the distributed loads indicated, determine the reaction at A.

Answer 1

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

                       

                   ∑ $F_y,net = 0 = R_A + P3 + P4 - P1 - P2$

                    $R_A = 4,500 + 1,734.375 - 984.375 -2,250\\\\R_A = 3,000 Lb$

      Moment Balance: Sum of moments about A = 0

      ∑ $M_n_e_t = 0 = M_A + P3*x3 + P4*x4 - P1*x1 - P2*x2$

                      $M_A = -(984.375)*(13.541667) - (2250)*(17.5) + (4500)*(3) + (1734.375)*\\(7.54166667)\\\\M_A = -26,125 Lb.ft$

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 ).