Chapter 10, Problem 10.4.21P

Uniform load q = 10 lb/ft acts over part of the span of fixed-end beam AB (see figure). Upward load P = 250 lb is applied 9 ft to the right of joint A. Find the reactions at A and B.

  

To determine

The reaction at fixed joint $A$ and $B$point.

Answer to Problem 10.4.21P

The reaction at $A$ is $-35.68\text{lbf}$ .

The moment at $A$is $261.6\text{lbf}\cdot \text{ft}$ .

The reaction at $B$ is $-154.32\text{lbf}$ .

The moment at $B$ is $-506.4\text{lbf}\cdot \text{ft}$ .

Explanation of Solution

Given information:

The length of the beam is $15\text{ft,}$ uniform load of $10\text{lb}/\text{ft}$ act over beam up to $6\text{ft,}$ and a upward load of $250\text{lb}$ is applied at $9\text{ft}$ from the fixed and A.

   Figure below shows the free body diagram of fixed beam.

  

   Figure-(1)

Write the expression for the reactions forces at point A.

  $R_{Ap}=\left(\frac{p{b}_p}{L}+\frac{M_A}{L}-\frac{M_B}{L}\right)$ …... (I)

Here, the point load is p, the length of the beam is L, the moment s acting at A and B is $M_{AP}$ and $M_p$ respectively.

Write the expression for the reactions forces at point $B$ .

  $R_{Bp}=\left(\frac{p{a}_p}{L}+\frac{M_{Ap}}{L}-\frac{M_{Bp}}{L}\right)$  .......(II)

Write the expression for the compatibility Equation at $B$ .

  $\begin{array}{c}{\theta }_A={\left({\theta }_A\right)}_1-{\left({\theta }_A\right)}_2-{\left({\theta }_A\right)}_3\\ =0\end{array}$  .......(III)

Here, the angles at the corresponding ends are represented by the subscript for $\theta$ .

Write the expression for the compatibility Equation at $B$ .

  $\begin{array}{c}{\theta }_B={\left({\theta }_B\right)}_1-{\left({\theta }_B\right)}_2-{\left({\theta }_B\right)}_3\\ =0\end{array}$  .......(IV)

Write the expression for the values of slopes of type 1 loading at point $A$.

  ${\left({\theta }_A\right)}_1=-\frac{p{a}_p{b}_p\left(L+b_p\right)}{6LEI}$

Write the expression for the value of slopes of type 1 loading at point $B$ .

  ${\left({\theta }_B\right)}_1=\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}$

Write the expression for the value of slopes for type 2 loading at point $A$ .

  ${\left({\theta }_A\right)}_2=-\frac{M_{AP}L}{3EI}$

Write the expression for the values of slopes for type 2 loading at point $B$ .

  ${\left({\theta }_B\right)}_2=-\frac{M_{AP}L}{6EI}$

Write the expression for the values of slopes for type 2 loading at point $A$${\left({\theta }_A\right)}_3=-\frac{M_{BP}L}{6EI}$

Write the expression for the values of slopes for type 2 loading at point $B$.

  ${\left({\theta }_B\right)}_4=-\frac{M_{BP}L}{3EI}$

Substitute $\frac{p{a}_p{b}_p\left(L+b_p\right)}{6LEI}$ for ${\left({\theta }_A\right)}_1$

  $-\frac{M_{AP}L}{3EI}$ for $$

  ${\left({\theta }_A\right)}_2$

  $\frac{M_{BP}L}{6EI}$ for ${\left({\theta }_A\right)}_3$ in Equation (III).

  $\begin{array}{l}\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}+\frac{M_{AP}L}{3EI}+\frac{M_{BP}L}{6EI}=0\\ \frac{M_{AP}L}{3EI}+\frac{M_{BP}L}{6EI}=-\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}\\ \end{array}$….. (V)

Substitute $\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}$ for ${\left({\theta }_B\right)}_1$

  $-\frac{M_{bP}L}{6EI}$ for ${\left({\theta }_B\right)}_2$

  $-\frac{M_{aP}L}{3EI}$ in Equation (IV).

  $\begin{array}{l}\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}+\frac{M_{AP}L}{6EI}+\frac{M_{BP}L}{3EI}=0\\ \frac{M_{AP}L}{6EI}+\frac{M_{BP}L}{3EI}=-\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}\end{array}$  .......(VI)

Substitute $\frac{M_{AP}L}{6EI}+\frac{M_{BP}L}{3EI}$ for $-\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}$ in equation (V) to obtain the values of $M_{AP}$ .

  $M_{AP}=-\frac{P{a}_p{b}_p^2}{L^2}$

Substitute $\frac{M_{AP}L}{6EI}+\frac{M_{BP}L}{3EI}$ for $-\frac{p{a}_p{b}_p\left(L+a_p\right)}{6LEI}$ in Equation (V).

  $M_{BP}=-\frac{P{a}_p{b}_p^2}{L^2}$

Substitute $-\frac{P{a}_p^2{b}_p}{L^2}$ for $M_A$ and $-\frac{P{a}_p{b}_p^2}{L^2}$ for $M_B$ in Equation (I).

  $\begin{array}{c}{R}_{AP}=\frac{p{b}_p}{L}-\frac{P{a}_p^2{b}_p}{L^2}+\frac{P{a}_p{b}_p^2}{L^2}\\ =-\frac{P{b}_p^2}{L^3}\left(L+2a_p\right)\end{array}$

Substitute $-\frac{P{a}_p{b}_p^2}{L^2}$ for $M_A$ and $-\frac{P{a}_p^2{b}_p}{L^2}$ for $M_B$ in Equation (II).

  $\begin{array}{c}{R}_{AP}=\frac{p{a}_p}{L}+\frac{P{a}_p{b}_p^2}{L^2}-\frac{P{a}_p^2{b}_p}{L^2}\\ =-\frac{P{a}_p^2}{L^3}\left(L+2b_p\right)\end{array}$

Write the expression for the fixed end moments due to the element of load $Ap$ .

  $d{M}_{AP}=\frac{qx{\left(L-x\right)}^{2}dx}{L^2}$

Write the expression for the fixed end moments due to the element of load $Bq$ .

  $d{M}_{Bq}=\frac{q{x}^2\left(L-x\right)dx}{L^2}$

Here the uniform distributed load is $q,$ the length of the beam is $L,$ and distance of the section is from the left side is $x$ .

Write the expression the moment at $A$.

  $M_{Aq}={\displaystyle \int d{M}_{Aq}}$  .......(VII)

Substitute $\frac{q{x}^2\left(L-x\right)dx}{L^2}$ for $d{M}_{Aq}$ in Equation (VII).

  $\begin{array}{c}{M}_{Aq}={\displaystyle \int \frac{q{x}^2\left(L-x\right)dx}{L^2}}\\ =\frac{q}{L^2}{\displaystyle {\int }_0^{a}x{\left(L-x\right)}^2}\\ =\frac{q{a}^2}{12L^2}\left(6L^2-8aL+3a^2\right)\end{array}$

Write the expression for the moment at point B.

  $M_{Bq}={\displaystyle \int d{M}_{Bq}}$  .......(VIII)

Substitute $\frac{q{x}^2\left(L-x\right)dx}{L^2}$ for $d{M}_B$ in Equation (VIII).

  $\begin{array}{c}{M}_{Bq}={\displaystyle \int \frac{q{x}^2\left(L-x\right)dx}{L^2}}\\ =\frac{q}{L^2}{\displaystyle \underset{0}{\overset{a}{\int }}{x}^2(L-x)dx}\\ =\frac{q{a}^3}{12L^2}\left(4L-3a\right)\end{array}$

Write the expression for the reaction force in term of elemental load $Aq$ .

  $d{R}_{Aq}=\frac{q{\left(L-x\right)}^2\left(L+2x\right)dx}{L^3}$

Write the expression for the reaction force in term of elemental load $Bp$ .

  $d{R}_{Bq}=\frac{q{x}^2\left(3L-2x\right)dx}{L^3}$

Write the expression for the reaction at point $A$ in term of element load.

  $R_{Aq}={\displaystyle \int d{R}_{Aq}}$….. (IX)

Substitute $\frac{q{\left(L-x\right)}^2\left(L+2x\right)dx}{L^3}$ for $d{R}_{Aq}$ in Equation (IX)

  $\begin{array}{c}d{R}_{Aq}={\displaystyle \int \frac{q{\left(L-x\right)}^2\left(L+2x\right)dx}{L^3}}\\ =\frac{q}{L^3}{\displaystyle \underset{0}{\overset{a}{\int }}{\left(L-x\right)}^2\left(L+2x\right)dx}\\ =\frac{qa}{2L^3}\left(2L^3-2L{a}^2+a^3\right)\end{array}$

Write the expression for the reaction at point $B$ in term of elemental load.

  $R_{Bq}={\displaystyle \int d{R}_{Bq}}$  .......(X)

Substitute $\frac{q{x}^2\left(3L-2x\right)dx}{L^3}$ for $d{R}_{Bq}$ in Equation (IX).

  $\begin{array}{c}{R}_{Bq}={\displaystyle \int \frac{q{x}^2\left(3L-2x\right)dx}{L^3}}\\ =\frac{q}{L^3}{\displaystyle {\int }_0^a{x}^2\left(3L-2x\right)dx}\\ =\frac{q{a}^3\left(2L-a\right)}{2L^3}\end{array}$

Write the expression for the reaction at point A.

  $R_A=R_{AP}+R_{Aq}$  .......(XI)

Substitute $-\frac{P{b}_p^2}{L^3}\left(L+2a_p\right)$ for $R_{AP}$ and $\frac{qa}{2L^3}\left(2L^3-2L{a}^2+a^3\right)$ for $R_{Aq}$ in Equation (XI).

  $R_A=-\frac{P{b}_p^2}{L^3}\left(L+2a_p\right)+\frac{qa}{2L^3}\left(2L^3-2L{a}^2+a^3\right)$  .......(XII)

Write the expression for the reactions at point B.

  $R_B=-\frac{P{a}_p^2}{L^3}\left(L+2b_p\right)+\frac{q{a}^3}{2L^3}\left(2L-a\right)$  .......(XIII)

Write the expression for the moment at point A.

  $M_A=M_{Ap}+M_{Aq}$  .......(XIV)

Substitute $-\frac{P{a}_p{b}_p{}^2}{L^2}$ for $M_{Ap}$ and $\frac{q{a}^2}{12L^2}\left(6L^2-8aL+3a^2\right)$ for $M_{Aq}$ in equation in Equation (XIII).

  $M_A=-\frac{P{a}_p{b}_p{}^2}{L^2}+\frac{q{a}^2}{12L^2}\left(6L^2-8aL+3a^2\right)$….. (XV)

Write the expression for the reaction moment at point B.

  $M_B=M_{Bp}+M_{Bq}$  .......(XVI)

Substitute $-\frac{P{a}_p{b}_p{}^2}{L^2}$ for $M_{Bp}$ and $\frac{q{a}^3}{12L^2}\left(4L-3a\right)$ for $M_{Bq}$ in Equation (XVI)

  $M_B=-\frac{P{a}_p{b}_p{}^2}{L^2}+\frac{q{a}^3}{12L^2}\left(4L-3a\right)$……. (XVII)

Calculation:

Substitute $250\text{lbf}$ for $p$ , $6\text{ft}$ for $b_p$ , $9\text{ft}$ for $a_p$ , $15\text{ft}$ for $L$ , $10\text{lbf}/\text{ft}$ for $q$ and $6\text{ft}$ for $a$ in Equation (XII).

  $\begin{array}{c}{R}_A=-\frac{\text{250lbf}\times {\left(\text{6}\text{ft}\right)}^{\text{2}}}{{\left(\text{15}\text{ft}\right)}^{\text{3}}}\left(\text{15}\text{ft}+\text{2}\left(\text{9}\text{ft}\right)\right)+\\ \left\{\frac{\text{10}\text{lbf}/\text{ft}\times \text{6}\text{ft}}{\text{2}\times {\left(\text{15}\text{ft}\right)}^{\text{3}}}\left(\begin{array}{l}\text{2}{\left(\text{15}\text{ft}\right)}^{\text{3}}\\ -\text{2}\left(\text{15}\text{ft}\right){\left(\text{6}\text{ft}\right)}^{\text{2}}\text{+}{\left(\text{6}\text{ft}\right)}^{\text{3}}\end{array}\right)\right\}\\ =\text{2}\text{.667}\text{lbf}/\text{ft×33ft}+\text{8}\text{.88}\times {\text{10}}^{\text{-3}}\left(\text{lbf}/{\text{ft}}^{\text{3}}\right)\times \text{5886}\left({\text{ft}}^{\text{3}}\right)\\ \text{=}-\text{88}\text{lbf}+\text{52}\text{.32}\text{lbf}\\ \text{=}-\text{35}\text{.68}\text{lbf}\end{array}$

Substitute $250\text{lbf}$ for p, $6\text{ft}$ for $b_p$ , $9\text{ft}$ for $a_p$ , $15\text{ft}$ for $L$ , $10\text{lbf}/\text{ft}$ for q and 6ft for $a$ in Equation (XIII).

  $\begin{array}{c}{R}_B=\left[\begin{array}{c}-\frac{250\text{lbf}\times {\left(9\text{ft}\right)}^2}{{\left(15\text{ft}\right)}^3}\left(15\text{ft}+\text{2}\left(6\text{ft}\right)\right)+\\ \left\{\frac{10\text{lbf}/\text{ft}\times {\left(6\text{ft}\right)}^3}{2\times {\left(15\text{ft}\right)}^3}\left(2\left(15\text{ft}\right)\right)-\left(6\text{ft}\right)\right\}\end{array}\right]\\ =-\left(6\text{lbf}/\text{ft}\right)\times \left(27\text{ft}\right)+\left(\text{0}\text{.32}\text{lbf}/\text{ft}\right)\times \left(\text{24}\text{ft}\right)\\ \text{=}-\text{162}\text{lbf}+\text{7}\text{.68}\text{lbf}\\ \text{=}-\text{154}\text{.32}\text{lbf}\end{array}$

Substitute $250\text{lbf}$for $p$ , $6\text{ft}$ for $b_p$

  $9\text{ft}$ for $a_p$ , $15\text{ft}$ for $L$

  $10\text{lbf}/\text{ft}$ for $q$ ,and 6ft for $a$ .

  $\begin{array}{l}{M}_A=-\frac{\text{250}\text{lbf}/\text{ft}\times \text{9}\text{ft}\times {\left(\text{6}\text{ft}\right)}^{\text{2}}}{\left(\text{15}{\text{ft}}^{\text{2}}\right)}+\\ \left[\frac{\text{10}\text{lbf}/\text{ft}\times {\left(\text{6}\text{ft}\right)}^{\text{2}}}{\text{12}\times {\left(\text{15}\text{ft}\right)}^{\text{2}}}\left(\begin{array}{l}\text{6}\text{ft}{\left(\text{15}\text{ft}\right)}^{\text{2}}-\\ \text{8}\left(\text{6}\text{ft}\right)\left(\text{15}\text{ft}\right)+\text{3}{\left(\text{6}\text{ft}\right)}^{\text{2}}\end{array}\right)\right]\\ \text{=}-\text{360}\text{lbf}\cdot \text{ft}+\text{98}\text{.4}\text{lbf}\cdot \text{ft}\\ \text{=}-\text{261}\text{.6}\text{lbf}\cdot \text{ft}\end{array}$

Substitute $250\text{lbf}$ for $p$ , $6\text{ft}$ for $b_p$

  $9\text{ft}$ for $a_p$ , $15\text{ft}$ for $L$

  $10\text{lbf}/\text{ft}$ for $q$ and $6\text{ft}$ for $a$ .

  $\begin{array}{c}{M}_B=-\frac{\text{250}\text{lbf}\times {\left(\text{9ft}\right)}^{\text{2}}\times \text{6}\text{ft}}{{\left(\text{15}\text{ft}\right)}^{\text{2}}}\\ +\left[\frac{\text{10}\text{lbf}/\text{ft}\times {\left(\text{6}\text{ft}\right)}^{\text{3}}}{\text{12}\times {\left(\text{15}\text{ft}\right)}^{\text{2}}}\left(\text{4}\left(\text{15}\text{ft}\right)-\text{3}\left(\text{6}\text{ft}\right)\right)\right]\\ \text{=}-\text{540}\text{lbf}\cdot \text{ft}+\text{33}\text{.6}\text{lbf}\cdot \text{ft}\\ \text{=}-\text{506}\text{.4l}\text{bf}\cdot \text{ft}\end{array}$

Conclusion:

The reaction at point $A$is$-35.68\text{lbf}$ .

The reaction at point $B$ is $-154.32\text{lbf}$ .

The moment at point $A$is$261.6\text{lbf}\cdot \text{ft}$ .

The moment at point $B$ is $-506.4\text{lbf}\cdot \text{ft}$ .