Question Question 1 10 points Save Answer Draw the influence line for vertical reaction at support B, shear at Cand the moment at point of the beams shown. From the influence line of moment at D, what is the value of moment Mo when x = 13 m due to a unit load? Use four decimal places. B -5 m -8 m -8 m -5 m-

Use Muller-Breslau's principle.Find the general shape of the influence line for vertical reaction \left(B_{y}\right) at supports B.Remove the roller support at B from the given beam and apply a small displacement \Delta. Due to the displacement, the structure releasedin which the deflected shape of the released structure of the beam is drawn.Draw the deflected shape of the beam without the roller support B as in Figure (1).The general shape of the influence line for vertical reaction \left(B_{y}\right) at supports B is the similar of deflected shape.Draw the general shape of the influence line for vertical reaction \left(B_{y}\right) at supports B using deflected shape as in Figure (2).Find the numerical value of the influence-line ordinate at B.Apply 1 \mathrm{kN} unit load at point B.Generally the influence line ordinate \left(B B^{\prime}\right) at B is 1 \mathrm{kN} / \mathrm{kN} when 1 \mathrm{kN} load is placed at B. The value of influence line ordinate in the different points on the beam determined by geometry (similar triangle) of the influence line.Find the influence line ordinate \left(C C^{\prime}\right) at C using similar triangle method.\begin{array}{l}\frac{A A^{\prime}}{A D}=\frac{B B^{\prime}}{B D} \\\frac{A A^{\prime}}{21}=\frac{1}{16} \\A A^{\prime}=\frac{21}{16}\end{array}Similarly calculate the influence line ordinate for B_{y} in different points on beam.Summarize the calculated values of influence line ordinate for B_{y} in different points on beam as in Table (1).Draw the influence line diagram for the vertical reactions at support B using Table 1 as in Figure (3).Remove the hinged support at D from the given beam and apply a small displacement \Delta. Due to the displacement, the structure releasedin which the deflected shape of the released structure of the beam is drawn.Draw the deflected shape of the released beam when hinged support D is removed as in Figure (4).The general shape of the influence line for vertical reaction \left(D_{y}\right) at supports D is the similar of deflected shape.Draw the general shape of the influence line for vertical reaction \left(D_{y}\right) at supports C using deflected shape as in Figure (5).Apply 1 \mathrm{kN} unit load at point D.Find the numerical value of the influence-line ordinate at D.Generally the influence line ordinate \left(D D^{\prime}\right) at D is 1 \mathrm{kN} / \mathrm{kN} when 1 \mathrm{kN} load is placed at D. The value of influence line ordinate in the different points on the beam determined by geometry (similar triangle) of the influence line.Refer Figure (5),Find the influence line ordinate \left(E E^{\prime}\right) at E using similar triangle method.\begin{aligned}\frac{E E^{\prime}}{B E} & =\frac{D D^{\prime}}{B D} \\\frac{E E^{\prime}}{21} & =\frac{1}{16} \\E E^{\prime} & =\frac{21}{16}\end{aligned}Similarly calculate the influence line ordinate for D_{y} in different points on beam.Summarize the calculated values of influence line ordinate for D_{y} in different points on beam as in Table (1).Draw the influence line diagram for the vertical reactions at support D as in Figure (6).Cut the beam at C to obtain the released structure of the beam. Next apply a small relative displacement in positive direction of S_{C} by moving end C of the portion A C downward by \Delta_{1} and end C of the portion C D upward by \Delta_{2} to obtain the deflected shape.Draw the deflected shape of the released beam when the beam cut at C as in Figure (7).The general shape of the influence line for shear \left(S_{C}\right) at C is the similar of deflected shape.Draw the general shape of the influence line for \operatorname{shear}\left(S_{C}\right) at C using deflected shape as in Figure (8).Find the numerical value of the influence-line ordinate at C.Apply 1 \mathrm{kN} unit load at first just to the left of C and then just to the right of C.Draw the free body diagram of beam as in Figure (9).Refer Figure ( 9),Consider clockwise moment as negative and anticlockwise moment as positive.Find the vertical reaction at B :Take moment at point B is equal to zero.\begin{array}{l}\Sigma M_{B}=0 \\\left(B_{y} \times 16\right)-(1 \times 8)=0 \\16 B_{y}=8 \\B_{y}=\frac{1}{2} \mathrm{kN}(\uparrow)\end{array}Consider upward force as positive.Find the vertical reaction at \mathrm{D} using vertical equilibrium equation:\begin{array}{l}\Sigma F_{y}=0 \\B_{y}-1+D_{y}=0\end{array}Substitute \frac{1}{2} \mathrm{kN} for B_{y}.\begin{array}{l}\frac{1}{2}-1+D_{y}=0 \\-\frac{1}{2}+D_{y}=0 \\D_{y}=\frac{1}{2} \mathrm{kN}\end{array}Find the shear \left(S_{C, L}\right) at C when the unit load is at just to the left of C.\begin{aligned}S_{C, L} & =-D_{y} \\& =-\frac{1}{2} \mathrm{kN}\end{aligned}Hence, the value of influence line ordinate just left of C is -\frac{1}{2} \mathrm{kN}.Hence, the value of influence line ordinate just left of C is -\frac{1}{2} \mathrm{kN}.Find the shear \left(S_{C, R}\right) when the unit load is at just to the right of C.\begin{aligned}S_{C, R} & =B_{y} \\& =\frac{1}{2} \mathrm{kN}\end{aligned}Hence, the value of influence line ordinate just right of C is -\frac{1}{2} \mathrm{kN}.The numerical value of influence line ordinate in different points on the beam for shear S_{C} at C is determined using geometry (similar triangle) of the influence line.Refer Figure (8),Find the influence line ordinate \left(A A^{\prime}\right) for S_{C} at A using similar triangle method.\begin{array}{l}\frac{A A^{\prime}}{A B}=\frac{C C^{\prime}}{B C} \\\frac{A A^{\prime}}{5}=\frac{\frac{1}{2}}{8} \\A A^{\prime}=\frac{5}{16}\end{array}Similarly calculate the influence line ordinate for S_{C} in different points on beam.Summarize the calculated values of influence line ordinate for S_{C} in different points on beam as in Table (3).Draw the influence line diagram for shear \left(S_{C}\right) at C using Table 3 as in Figure (10).Apply a hinge at C and to obtain deflected shape and released structure of the beam. The general shape of the influence line for the moment \left(M_{C}\right) at point C is the similar of deflected shape.Draw the deflected shape of the released beam when insert a hinge at C as in Figure (11).The general shape of the influence line for bending moment \left(M_{C}\right) at C is the similar of deflected shape.Draw the general shape of the influence line for \operatorname{moment}\left(M_{C}\right) at C using deflected shape as in Figure (12).Refer Figure ( 9 ),Consider clockwise moment as negative and anticlockwise moment as positive.Find the numerical value of the influence-line ordinate for M_{C} at C.Take moment equilibrium at point C.M_{C}=B_{y}(8)Substitute \frac{1}{2} \mathrm{kN} for B_{y}.\begin{aligned}M_{C} & =\frac{1}{2}(8) \\& =4 \mathrm{kN}\end{aligned}Hence, the influence line ordinate of moment at C is 4 \mathrm{kN}-\mathrm{m} / \mathrm{kN}.The numerical value of influence line ordinate in different points on the beam for moment M_{C} is determined using geometry (similar triangle) of the influence lineRefer Figure (8),Find the influence line ordinate \left(A A^{\prime}\right) for M_{C} at A using similar triangle method.\begin{array}{l}\frac{-A A^{\prime}}{A B}=\frac{C C^{\prime}}{B C} \\\frac{A A^{\prime}}{5}=-\frac{4}{8} \\A A^{\prime}=-\frac{5}{2}\end{array}Similarly calculate the influence line ordinate for M_{C} in different points on beam.Summarize the calculated values of influence line ordinate for M_{C} in different points on beam as in Table (4).Draw the influence line diagram for moment \left(M_{C}\right) at C using Table 4 as in Figure (13).Hence, the influence lines for vertical reaction at B and \mathrm{D} and the shear and moment at point \mathrm{C} of the beam are drawn. ...