# Problem 736 | Shear and moment diagrams of fully restrained beam under triangular load

**Problem 736**

Determine the end shears and end moments for the restrained beam shown in Fig. P-736 and sketch the shear and moment diagrams.

**Solution 736**

$\frac{1}{2}(12)(12R_A) + 12M_A - \frac{1}{4}(8)(5760) = 0$

$72R_A + 12M_A - 11\,520 = 0$

$6R_A + M_A = 960$ → Equation (1)

$EI\,t_{C/A} = (Area_{AC}) \cdot \bar{X}_C = 0$

$\frac{1}{2}(12)(12R_A)[ \, \frac{1}{3}(12) \, ] + 12M_A[ \, \frac{1}{2}(12) \, ] - \frac{1}{4}(8)(5760)[ \, \frac{1}{5}(8) \, ] = 0$

$288R_A + 72M_A - 18\,432 = 0$

$4R_A + M_A = 256$ → Equation (2)

Solving Equations (1) and (2)

$R_A = 352 \, \text{ lb}$ *answer*

$M_A = -1152 \, \text{ lb}\cdot\text{ft}$ *answer*

**Checking**

$EI\,t_{A/C} = (Area_{AC}) \cdot \bar{X}_A = 0$

$\frac{1}{2}(12)(12R_A)[ \, \frac{2}{3}(12) \, ] + 12M_A[ \, \frac{1}{2}(12) \, ] - \frac{1}{4}(8)(5760)[ \, 4 + \frac{4}{5}(8) \, ] = 0$

$576R_A + 72M_A - 119\,808 = 0$

$576(352) + 72(-1152) - 119 808 = 0$

$0 = 0$ *okay!*

$M_C = 12R_A + M_A - 5760$

$M_C = 12(352) - 1152 - 5760$

$M_C = -2688 \, \text{ lb}\cdot\text{ft}$ *answer*

$\Sigma F_V = 0$

$R_A + R_C = \frac{1}{2}(8)(540)$

$352 + R_C = 2160$

$R_C = 1\,808 \, \text{ lb}$ *answer*

**To draw the shear diagram**

- V
_{A}= 352 lb - V
_{B}= V_{A}+ Load_{AB}

V_{B}= 352 + 0

V_{B}= 352 lb - There is no load between AB, thus, shear in AB is constant.
- V
_{C}= V_{B}+ Load_{BC}

V_{C}= 352 - (1/2)(8)(540)

V_{C}= -1808 lb - Load between B and C is linearly decreasing from zero to -540 lb/ft, thus, shear in segment BC is a concave downward second degree curve (parabola) with vertex at B.
- Location of point D by squared property of parabola:

$\dfrac{{x_D}^2}{352} = \dfrac{8^2}{352 1808}$$x_D = 3.23 \, \text{ ft}$ to the right of B

**To draw the moment diagram**

- M
_{A}= -1152 lb·ft - M
_{B}= M_{A}+ V_{AB}

M_{B}= -1152 + 352(4)

M_{B}= 256 lblb·ftft - The shear between A and B is uniform and positive, thus, the moment in AB is linear and increasing.
- M
_{D}= M_{B}+ V_{BD}

M_{D}= 256 + (2/3)(x_{D})(352)

M_{D}= 256 + (2/3)(3.23)(352)

M_{D}= 1013.97 lb·ft - M
_{C}= M_{D}+ V_{DC}

Solving for VDC

V_{DC}= (-1/3)(8)(352 + 1808) + (1/3)(x_{D})(352) + 352(8 - x_{D})

V_{DC}= -5760 + (1/3)(3.23)(352) + 352(8 - 3.23)

V_{DC}= -3701.97 lb

Thus,

M_{C}= 1013.97 - 3701.97

M_{C}= -2688 lb·ft - The shear diagram from B to C is a parabola, thus, the moment diagram of segment BC is a third degree curve. The value of shear from B to C decreases, thus, the slope of moment diagram between B and C also decreases making the cubic curve concave downward.

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