Question

24 investigation of forces, force systems, loading, and reactions 2.22 kn1.78 kn 2.67 kn 500 lb 400 lb 600 lb 1.2 m 0.6 1.2 m 1.8 m 850 lb 650 lb 3.78 kn 2.89 kn figure 1.18 reference for problem 1.8.a. the reader should verify this case by selecting any other point, such as point c in figure 1.17c as the center of moments, and confirming that the sum of the moments is zero for this point. problem 1.8.a figure 1.18 represents a beam in equilibrium with three loads and two reactions. select five different centers of moments and write the equation of moments for each, showing that the sum of the clockwise moments equals the sum of the counterclockwise moments.

Explanation:

Step1: Recall moment formula

$M = F\times d$, where $M$ is moment, $F$ is force and $d$ is perpendicular - distance from the point of rotation to the line of action of the force. Clock - wise moments are considered positive and counter - clockwise moments are negative.

Step2: Select first center of moments (left - hand support)

Let the left - hand support be the center of moments. The 500 lb force creates a clock - wise moment $M_1=500\times4$. The 400 lb force creates a clock - wise moment $M_2 = 400\times(4 + 2)$. The 600 lb force creates a clock - wise moment $M_3=600\times(4 + 2+4)$. The right - hand reaction of 650 lb creates a counter - clockwise moment $M_4=-650\times(4 + 2+4 + 6)$. The sum of moments $\sum M_1=500\times4+400\times(4 + 2)+600\times(4 + 2+4)-650\times(4 + 2+4 + 6)$.

Step3: Select second center of moments (at 500 lb load)

Let the point of application of the 500 lb load be the center of moments. The 400 lb force creates a clock - wise moment $M_5 = 400\times2$. The 600 lb force creates a clock - wise moment $M_6=600\times(2 + 4)$. The left - hand reaction of 850 lb creates a counter - clockwise moment $M_7=-850\times4$. The right - hand reaction of 650 lb creates a counter - clockwise moment $M_8=-650\times(4 + 2+4)$. The sum of moments $\sum M_2=400\times2+600\times(2 + 4)-850\times4-650\times(4 + 2+4)$.

Step4: Select third center of moments (at 400 lb load)

Let the point of application of the 400 lb load be the center of moments. The 500 lb force creates a counter - clockwise moment $M_9=-500\times2$. The 600 lb force creates a clock - wise moment $M_{10}=600\times4$. The left - hand reaction of 850 lb creates a counter - clockwise moment $M_{11}=-850\times(4 + 2)$. The right - hand reaction of 650 lb creates a counter - clockwise moment $M_{12}=-650\times(2 + 4)$. The sum of moments $\sum M_3=-500\times2+600\times4-850\times(4 + 2)-650\times(2 + 4)$.

Step5: Select fourth center of moments (at 600 lb load)

Let the point of application of the 600 lb load be the center of moments. The 500 lb force creates a counter - clockwise moment $M_{13}=-500\times(4 + 2)$. The 400 lb force creates a counter - clockwise moment $M_{14}=-400\times4$. The left - hand reaction of 850 lb creates a counter - clockwise moment $M_{15}=-850\times(4 + 2+4)$. The right - hand reaction of 650 lb creates a counter - clockwise moment $M_{16}=-650\times6$. The sum of moments $\sum M_4=-500\times(4 + 2)-400\times4-850\times(4 + 2+4)-650\times6$.

Step6: Select fifth center of moments (right - hand support)

Let the right - hand support be the center of moments. The 500 lb force creates a counter - clockwise moment $M_{17}=-500\times(4 + 2+4 + 6)$. The 400 lb force creates a counter - clockwise moment $M_{18}=-400\times(2+4 + 6)$. The 600 lb force creates a counter - clockwise moment $M_{19}=-600\times(4 + 6)$. The left - hand reaction of 850 lb creates a clock - wise moment $M_{20}=850\times(4 + 2+4 + 6)$. The sum of moments $\sum M_5=-500\times(4 + 2+4 + 6)-400\times(2+4 + 6)-600\times(4 + 6)+850\times(4 + 2+4 + 6)$.

Answer:

The equations for the sum of moments for each of the five centers of moments are calculated as shown in the steps above. For a beam in equilibrium, in each case, the sum of clock - wise moments should equal the sum of counter - clockwise moments.