# Solution to Problem 587 | Design for Flexure and Shear

**Problem 587**

A beam carries two concentrated loads P and triangular load of 3P as shown in Fig. P-587. The beam section is the same as that in Fig. P-577 on this page. Determine the safe value of P if f_{b} ≤ 1200 psi and f_{v} ≤ 200 psi.

**Solution 587**

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

$R_1 = 2P$

$\Sigma M_{R1} = 0$

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

$R_2 = 3P$

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

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

**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*