7-53 Draw the Shear and Bending-moment Diagrams for the Beam

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Statics- Example
Beam Loading and Supports		Example
		Draw the shear and moment diagram for the beam shown in effigy.
		Solution
Beam Complimentary-body Diagram		Earlier the shear and moment tin can be adamant at whatever internal location, the boundary weather need to be calculated. At that place will be three possible reactions at A and B namely Ax, Ay and By. To summate them, the distributed load is converted into an equivalent point load of (3 kN/m) (3 yard) = ix kN which is located three m from support A as shown in the diagram. The unknown reactions can exist institute by applying the three standard equilibrium equations. Taking moments about support A and assuming counter clockwise (CCW) moment equally positive gives,      ΣMA = 0 By (vi) - (5) (three) - (ix) (iii) = 0 By = 7 kN Equating vertical forces gives,      ΣFy = 0 Ay + By - five - 9 = 0 Ay = 7 kN Since there is no horizontal applied load, Ax volition be zero. (Reaction forces can as well exist calculated by dividing sum of forces acting on the beam by 2, since the beam and loading are symmetrical.)
Beam Sections		Since the beam has forces acting at four locations, the shear and moment needs to be analyzed in iv dissimilar sections. The shear and moment may not be continuous beyond a load or support. Each department needs to exist cutting and analyzed for shear and moment as a part of x from the left edge. The results can then be plotted and the maximum shear and moment can be easily identified.
		Section 1
Department i		To find the shear and moment in section 1, cut the beam at an arbitrary point x in department i and and then draw a free-torso diagram as shown on the left. Next, sum the forces to give,      ΣFy = 0 7 - 51 = 0 Fiveone = 7 kN Summing the moments almost the cut edge gives,      ΣMcutting = 0 M1 - 7x = 0 One thousand1 = 7x kN-g It should be noted that these results are only good for department 1 (0 ≤ x ≤ 1.5 thousand).
		Section 2
Section ii		Section two is between the start of distributed load and the point load. Over again, cut the beam at an arbitrary point x in section 2 then describe a gratis-body diagram as shown on the left. Sum the forces to give,      ΣFy = 0 7 - 3 (ten - 1.5) - V2 = 0 Vii = xi.five - 3x kN Summing the moments about the cut edge gives,      ΣMcut = 0 M2 + iii(ten - one.v)(ten - 1.5) / 2 - 7x = 0 M2 = 7x - i.v(x - 1.five)2 kN-m These results are only good for department 2 (1.5 ≤ ten ≤ 3m). To aid in graphing the shear and moment equations, the exact values can be calculated at each finish of the axle section. When x = i.five, the shear force volition be, Five2 = xi.five - (iii)1.five = 7 kN and the moment volition be, Yardthree = (vii)ane.5 - 1.5(1.5 - ane.5)ii = 10.5 kN-m When ten = three, the shear force will exist, V2 = xi.5 - (3)3 = 2.5 kN and the moment will exist, M3 = (seven)3 - 1.5(iii - 1.5)ii = 17.625 kN-m
		Section 3
Department 3		Section 3 is between point load of 5 kN and end of universal distributed load. The free-body diagram for the beam volition be as shown on the left. Sum of the forces gives,      ΣFy = 0 seven - 5 - iii(x - one.5) - Viii = 0 V3 = half dozen.5 - (3)ten kN Summing the moments almost the cutting border gives,      ΣMcutting = 0 Thou3 - (7)x + v(x - three) + iii(x - one.5)(10 - 1.v) / 2 = 0 M3 = 2x + 15 - 1.five(x - 1.5)2 kN-k These results are but skilful for section iii (iii ≤ x ≤ iv.5m). When x = three, the shear force will be, Fivethree = 6.5 - (3)three = -2.5 kN the moment will be, M3 = (2)three + 15 - i.5(iii - 1.5)2 = 17.625 kN-thousand When 10 = 4.five, the shear force volition be, V3 = 6.5 - (3)4.5 = -7 kN the moment will be, M3 = (ii)4.5 + 15 - 1.5(four.5 - 1.5)2 = x.v kN-m
		Section 4
Section 4		The final cut is at the far right department of the beam. The free-body diagram for the beam will be as shown on the left. Sum of the forces gives,      ΣFy = 0 vii - 5 - three(3) - 5four = 0 54 = -7 kN Summing the moments nearly the cut border gives,      ΣMcut = 0 Mfour - (seven)x + five(x - three) + three(3)(x - 3) = 0 M4 = (7)10 - 14(x-three) kN-m These results are only skillful for section 3 (four.5 ≤ x < 6m). When x = 4.v, the shear force volition be, V4 = -seven kN the moment will be, K4 = (seven)4.5 - 14(4.5 - 3 ) = 10.v kN-m When x = six, the shear force volition be, Five4 = -seven kN the moment will be, 10004 = (vii)6 - 14(6 - 3) = 0 kN-grand
		Shear and Moment Diagrams
		At present that the shear and moment is known for each section of the axle, the results can be plotted. The resulting graphics are called the shear diagram and moment diagram. Since the same ten was used for all three sections, the each equation for each department can exist hands plotted every bit shown at the left. Since beam and loading are symmetrical, shear force diagram and bending moment diagram tin can also be drawn by solving commencement ii sections simply. For shear moment diagram other one-half will be anti-mirror image of get-go 2 sections and for bending moment diagram other one-half will be mirror paradigm of commencement 2 sections which can exist seen in the figure shown on the left.

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7-53 Draw the Shear and Bending-moment Diagrams for the Beam

Source: https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=st&chap_sec=08.2&page=example

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