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ENGR20004_2025_SM2 Topic quiz - internal loads of beams

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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

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We start by restating the problem setup and the specific blanks to fill in, then evaluate each option for each blank. First dropdown (the Fy equation): Sum of vertical forces about the chosen axis is set to zero. The given expression is Sum Fy = 0 = (3 kN) − ½(2/3 x) x + [Select] (the [Select] is the first dropdown). - The term (3 kN) represents a vertical force shown in the figure, acting upward at the left end. The triangular distributed load up to section x has resultant equal to the area of the triangle, which is ½ × base × height = ½ × x × (2/3 x) = (1/3) x^2 kN, and it acts downward. - To satisfy equilibrium, the remaining unknown force at the cut or at the other end must provide the necessary vertical force balance. Since the distributed load is downward and the left end force is upward, the reaction term that completes the balance should be added with a positive sign in the Fy equation if V is defined as an upward reaction on the cut section (as is common in beam cut analyses). - Therefore, the correct sign for the f......Login to view full explanation

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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 )   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

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 Notes

When the shear force diagram passes through the zero axis, the corresponding bending moment at the same location will be...?

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