Problem 6.71 Consider the frame shown in Ugue 1). Suppose that Fi 16 kN Figure 1

Question

Problem 6.71 Consider the frame shown in Ugue 1). Suppose that Fi 16 kN Figure 1 3 kN/ m 6 m. 3 m 4 m 3m 211m. Part A Determine the rand y components of the reaction al the support A using scalar nolalion. Express your answers using three significant figures separated by a comma. AL A Submit My Answers Give Up Part B Determine the moment of reaction at A Express your answer to three significant figures and include the appropriate units. Enter positive value if the moment is counterclockwise and negative value if the moment is clockwise MA Value Units Submi My Answers Give U Part C Determine the y compon of the reaction at the support Eusing scalar notation. ent Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Let's take x component of reaction at A as Ax and y component as Ay and take the moment at A as Ma

similary at C as Cy and at E as Ey (at C and E there will be no x component as they are roller)

(Part A)

total force acting on rod = 16kN +3kN/m*4m = 28kN

this force will be balanced by all the reactions in vertical reaction

so Ay + Cy + Ey = 28 kN.......................(1)

as there is no force in horizontal direction

so Ax = 0

now see there is hinge at point B and D ,we know that there is no moment at hinge only vertical and horizontal force

now take only DE part of rod

there will be Dx and Dy forces at D and F1 = 16kN at the end and Ey as reaction

so Dy + Ey = 16 and Dx = 0

now take moment about D

so 16*9 - Ey*6 = 0 as there is no moment in DE part of rod

So Ey = 24 kN (j)

so Dy + Ey = 16 kN

so Dy = 8kN (-j) as in the negative y direction

Now consider Part BD of the rod which is also a hinged part of rod

forces on BD will be 3*4 =12kN (distributive force) , Cy as reaction , Dy at D and By at B. Here Bx will be zero again

now Since Dy was acting in the negative direction in DE part of road it will be acting in y direction here because of the balancing act, if were to combine the both end resultant should be zero

so write equation balancing forces we get

Cy + Dy + By = 12kN

now take Moment About B

Cy*4 + Dy*6 - 12*3 = 0 (here since it is a distributive force we can solve it as taking a net force acting at the very middle of the rod)

Cy*4 + 8*6 - 12*3=0

so Cy = 3kN (j) in positive y direction

now Cy + Dy + By = 12kN

so We get By = 1 kN (j) in positive y direction

now consider Part AB of rod

Here forces are Ax= 0, Ay , Ma and By

now Ay +By = 0 (here direction of By will be negative direction)

Ay = 1 kN (j) in positive y direction

now take moment about A

Ma+ 1*3 = 0

Ma = 3kNm

so Ax = 0 and Ay = 1 kN (j)

Ma = 3 kNm

Ey = 24 kN (j)

Cy = 3 kN (j)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---