Question

1. Calculate the centroid of the plane region shown. 1 in. 3 in. - 3 in.--e-3 in. - 2. Determine the reactions at the supports of the beam below. 3 kip/ft 2 kip/ft TTTIIIIIIIIIII Во с –6 ft- - 6 ft - 3ft-

Answer 1

1. The figure is divided into four shapes

Area of triangle = A₁ = (1/2)x3x3 = 4.5 in²

Distance of x-coordinate of centroid of triangle from extreme left = (2/3)x3 = x₁ = 2 in

Distance of y-cordinate of centroid of triangle from Base = y₁ = 3/3 = 1 in

Area of rectangle = A₂= 3x3 = 9 in²

x₂ = 3 + 1.5 = 4.5 in

y₂ = 3/2 = 1.5 in

Area of Quarter circle = A₃ = piex3²/4 = 7.068 in²

$x_{3}=3+3+\frac{4R}{3\pi }$

  $=6+\frac{4(3)}{3\pi }$

= 7.273 in

$y_{3}=\frac{4R}{3\pi }=\frac{4(3)}{3\pi }$

= 1.273 in

Area of semicircle = A₄ = piex1²/2

= 1.571 in

x₄ = 3 + 3 = 6 in

y₄ = 4R/3pie = 4x1/3xpie

= 0.424 in

X-coordinate of centroid of figure from extreme left is given as

$x'=\frac{A_{1}x_{1}+A_{2}x_{2}+A_{3}x_{3}-A_{4}x_{4}}{A_{1}+A_{2}+A_{_{3}}-A_{4}}$

= (4.5x2 + 9x4 + 7.068x7.273 - 1.571x6)/(4.5 + 9 + 7.068 - 1.571)

= 4.815 in

Y - coordinate of centroid of figure from base is given as

$y'=\frac{A_{1}y_{1}+A_{2}y_{2}+A_{3}y_{3}-A_{4}y_{4}}{A_{1}+A_{2}+A_{_{3}}-A_{4}}$

= 1.386 in

2. Let V_a and V_b be the vertical reactions at A and B respectively in the upwards direction and H_a be the horizontal reaction at A

Taking moments about B

Clockwise as +ve and Anticlockwise as -ve

$\sum M_{b}=0$

V_ax6 - (1/2)x3x6x(6/3) + 2x3x3/2 = 0

6V_a = 9

=> V_a = 1.5 kip (upwards)

For vertical equilibrium

V_a + V_b = (1/2)x3x6 + 2x3

=> V_b = 13.5 kip (upwards)

For horizontal equilibrium

H_a = 0