$\Sigma M_{R2} = 0$

$12R_1 + 4P = 16P + 4(3P)$

$R_1 = 2P$

$\Sigma M_{R1} = 0$

$12R_2 + 4P = 16P + 8(3P)$

$R_2 = 3P$

$W = \frac{1}{2}(12)w_o = 3P$

$w_o = \frac{1}{2}P$

**To draw the Shear Diagram**

- V
_{A} = -P lb
- V
_{B} = V_{A} + Area in load diagram

V_{B} = -P + 0 = -P lb

V_{B2} = V_{B} + R_{1} = -P + 2P = P lb
- V
_{C} = V_{B2} + Area in load diagram

V_{C} = P - ½(12)(½P) = -2P lb

V_{C2} = V_{C} + R_{2} = -2P + 3P = P lb
- V
_{D} = V_{C2} + Area in load diagram

V_{D} = P + 0 = P

V_{D2} = V_{D} - P = P - P = 0
- Shear at AB and CD are rectangular.
- Shear at BC is parabolic (2nd degree curve).
- Location of zero shear:

By squared property of parabola

x^{2} / P = 122 / 3P

x = 6.93 ft

12 - x = 5.07 ft

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - 4P = -4P lb·ft
- M
_{E} = M_{B} + Area in shear diagram

M_{E} = -4P + 2/3 (6.93)(P) = 0.62P lb·ft
- M
_{C} = M_{E} + Area in shear diagram

M_{C} = 0.62P - [ 1/3 (12)3P - 2.31P - 5.07P ]

M_{C} = -4P lb·ft
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = -4P + 4P = 0
- The moment diagram at AB and CD are straight lines (1st degree curves) while at BC is 3rd degree curve.

Based on allowable bending stress

$f_b = \dfrac{Mc}{I}$

Where (From Solution 577)

c = 6 in

I = 350.67 in^{4}

Thus,

$1200 = \dfrac{4P(12)(6)}{350.67}$

$P = 1461.125 \, \text{ lb}$

Based on allowable shear stress

$f_v = \dfrac{VQ}{Ib}$

Where (From Solution 577)

Q = 35.5 in^{3}

I = 350.67 in^{4}

b = 0.75 in

Thus,

$200 = \dfrac{2P(35.5)}{350.67(0.75)}$

$P = 740.85 \, \text{ lb}$

For safe value of P, use **P = 740.85 lb**. *answer*