Solution to Problem 456 | Moving Loads

$R = P_1 + P_2+ P_3$

$R = 4^k + 8^k + 6^k$

$R = 18 \, \text{kips}$

$R = 18,000 \, \text{lbs}$
 

 

$xR = 9P_2 + (9 + 18)P_3$

$x(18) = 9(8) + (9 + 18)(6)$

$x = 13 \, \text{ft} \,\,$ the resultant R is 13 ft from P₁
 

Maximum moment under P₁
$\Sigma M_{R2} = 0$

$44R_1 = 15.5R$

$44R_1 = 15.5(18)$

$R_1 = 6.34091 \, \text{kips}$

$R_1 = 6,340.91 \, \text{lbs}$
 

 

$M_{To\,\,the\,\,left\,\,of\,\,P_1} = 15.5R_1$

$M_{To\,\,the\,\,left\,\,of\,\,P_1} = 15.5(6340.91)$

$M_{To\,\,the\,\,left\,\,of\,\,P_1} = 98,284.1 \, \text{lb}\cdot\text{ft}$
 

Maximum moment under P₂
$\Sigma M_{R2} = 0$

$44R_1 = 20R$

$44R_1 = 20(18)$

$R_1 = 8.18182 \, \text{kips}$

$R_1 = 8,181.82 \, \text{lbs}$
 

 

$M_{To\,\,the\,\,left\,\,of\,\,P_2} = 20R_1 - 9P_1$

$M_{To\,\,the\,\,left\,\,of\,\,P_2} = 20(8181.82) - 9(4000)$

$M_{To\,\,the\,\,left\,\,of\,\,P_2} = 127,636.4 \, \text{lb}\cdot\text{ft}$
 

Maximum moment under P₃
$\Sigma R_1 = 0$

$44R_2 = 15R$

$44R_2 = 15(18)$

$R_2 = 6.13636 \, \text{kips}$

$R_2 = 6,136.36 \, \text{lbs}$
 

 

$M_{To\,\,the\,\,right\,\,of\,\,P_3} = 15R_2$

$M_{To\,\,the\,\,right\,\,of\,\,P_3} = 15(6,136.36)$

$M_{To\,\,the\,\,right\,\,of\,\,P_3} = 92,045.4 \, \text{lb}\cdot\text{ft}$
 

Thus,
$M_{max} = M_{To\,\,the\,\,left\,\,of\,\,P_2}$

$M_{max} = 127,636.4 \, \text{lb}\cdot\text{ft} \,\,$ answer
 

The maximum shear will occur when P₁ is over the support.
$\Sigma M_{R2} = 0$

$44R_1 = 31R$

$44R_1 = 31(18)$

$R_1 = 12.6818 \, \text{kips}$

$R_1 = 12,681.8 \, \text{lbs}$
 

 

Thus, $V_{max} = 12,681.8 \, \text{lbs} \,\,$ answer