Question text 10Marks The beam ABCDE shown in Figure P10 has two different uniformly distributed loads (UDL) acting, as well as an applied moment at A. The general shape of the shear force diagram (SFD) and the bending moment diagram (BMD) are also shown in the figure. Figure P10Use the following positive sign convention for the internal forces in the beam: (a) If: M = 11 kNm; RB = 15.887 kN; and, RD = 10.413 kN, Determine the value of the UDL q as shown on the figure? Provide your answer as a positive value in units of kN/m to 2 decimal places. q = Answer 1[input] kN/m (2 marks) Note that RB and RD are the values of the reactions at supports B and D, respectively, as shown in the figure, with positive values indicating vertically upwards reaction forces. Use the following data for sections (b) to (e). Now take that M = 13.8 kNm, and q = 9 kN/m. In this case, the reaction forces at supports B and D are RB = 16.87 kN and RD = 7.83 kN, respectively, with positive values indicating vertically upwards reaction forces. (b) What is the distance, labelled as ‘xm’, between B and C in the figure (measured from B), where the shear force value crosses zero on the shear force diagram? Provide your answer in meters (m) to 3 decimal places. xm = Answer 2[input] m  (2 marks) (c) What is the value of the shear force VC shown on the figure? Provide your answer in units of kN to 2 decimal places. Shear force at C, VC = Answer 3[input] kN (2 marks) (d) What is the value (including the sign) of the internal bending moment at point B, MB shown in the figure? Provide your answer in units of kNm to 2 decimal places. Internal Moment at B, MB = Answer 4[input] kNm (2 marks) (e) What is the value (including the sign) of the internal bending moment at point D, MD shown in the figure? Provide your answer in units of kNm to 2 decimal places. Internal Moment at D, MD = Answer 5[input] kNm (2 marks) Notes Report question issue Question 10 NotesMultiple fill-in-the-blank

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Step 2: Shear and Moment Functions A free-body diagram of a beam segment of length x is shown. The intensity of the triangular load at the section is found by proportion, that is, w/x=(2kN/m)/(3m) or   w=( 2 3  x)  . The resultant of the distributed loading is found from the area under the diagram. Thus, \sum F_y=0=(3\,kN)-\frac{1}{2}\left(\frac{2}{3}x\right)x [ Select ] +V -V +M -M V=\left(3-\frac{1}{3}x^2\right)kN \sum M_x=0=(6kNm)-(3kN)(x)+\frac{1}{2}(2/3\,x)x(1/3\,x) [ Select ] -Vx +Vx +M -M M=(-6+3x-\frac{1}{9}x^3) kNm   Choose the missing terms in the above equations

Step 2: Shear and Moment Functions A free-body diagram of a beam segment of length x is shown. The intensity of the triangular load at the section is found by proportion, that is, w/x=(2kN/m)/(3m) or   𝑤 = ( 2 3   𝑥 )   kN / m . The resultant of the distributed loading is found from the area under the diagram. Thus, \sum F_y=0=(3\,kN)-\frac{1}{2}\left(\frac{2}{3}x\right)x [ Select ] +V -V +M -M V=\left(3-\frac{1}{3}x^2\right)kN \sum M_x=0=(6kNm)-(3kN)(x)+\frac{1}{2}(2/3\,x)x(1/3\,x) [ Select ] -Vx +Vx +M -M M=(-6+3x-\frac{1}{9}x^3) kNm   Choose the missing terms in the above equations

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