7-55 Draw Shear and Moment Diagrams

Chapter

1. Basics

2. Vectors

iii. Forces

4. Moments

5. Rigid Bodies

6. Structures

vii. Centroids/Inertia

8. Internal Loads

9. Friction

10. Work & Energy

Appendix

Basic Math

Units

Sections

Search

eBooks

Dynamics

Statics

Mechanics

Fluids

Thermodynamics

Math

Author(s):

Kurt Gramoll

©Kurt Gramoll

Statics- Example
Beam Loading and Supports		Example
		Draw the shear and moment diagram for the beam shown in figure.
		Solution
Beam Free-body Diagram		Before the shear and moment tin be determined at any internal location, the purlieus weather need to exist calculated. There will be iii 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 m) = 9 kN which is located 3 m from support A as shown in the diagram. The unknown reactions can be constitute by applying the iii standard equilibrium equations. Taking moments about support A and bold counter clockwise (CCW) moment as positive gives,      ΣMA = 0 By (half dozen) - (5) (3) - (9) (3) = 0 By = 7 kN Equating vertical forces gives,      ΣFy = 0 Ay + By - five - 9 = 0 Ay = 7 kN Since there is no horizontal practical load, Ax will be nothing. (Reaction forces can as well be 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 iv locations, the shear and moment needs to be analyzed in four dissimilar sections. The shear and moment may non be continuous across a load or support. Each section needs to be cutting and analyzed for shear and moment equally a function of 10 from the left edge. The results tin can then be plotted and the maximum shear and moment tin can exist easily identified.
		Department 1
Section 1		To notice the shear and moment in section ane, cut the beam at an arbitrary indicate x in section i and then draw a free-torso diagram as shown on the left. Adjacent, sum the forces to give,      ΣFy = 0 7 - Five1 = 0 Fivei = 7 kN Summing the moments most the cut edge gives,      ΣMcut = 0 G1 - 7x = 0 Yardane = 7x kN-m It should be noted that these results are only skilful for department i (0 ≤ x ≤ 1.5 one thousand).
		Section ii
Section 2		Section two is between the start of distributed load and the point load. Again, cut the axle at an arbitrary point 10 in section 2 and then draw a free-body diagram as shown on the left. Sum the forces to requite,      ΣFy = 0 7 - iii (x - 1.5) - V2 = 0 V2 = 11.five - 3x kN Summing the moments about the cut edge gives,      ΣMcut = 0 M2 + three(ten - 1.5)(x - 1.5) / 2 - 7x = 0 Mtwo = 7x - 1.5(x - 1.5)two kN-m These results are but good for section 2 (i.v ≤ x ≤ 3m). To assist in graphing the shear and moment equations, the exact values can be calculated at each terminate of the beam section. When x = 1.five, the shear strength will be, Five2 = eleven.5 - (3)one.5 = 7 kN and the moment volition exist, G3 = (7)ane.5 - 1.5(1.5 - one.5)two = 10.5 kN-m When x = 3, the shear force will be, V2 = 11.5 - (three)three = ii.5 kN and the moment will be, Thousand3 = (7)3 - 1.5(three - 1.5)2 = 17.625 kN-m
		Section iii
Department 3		Section 3 is between point load of 5 kN and end of universal distributed load. The free-torso diagram for the beam will be equally shown on the left. Sum of the forces gives,      ΣFy = 0 vii - v - 3(x - ane.5) - V3 = 0 5three = six.five - (3)x kN Summing the moments near the cut edge gives,      ΣMcut = 0 Gthree - (7)x + v(10 - 3) + iii(ten - 1.5)(ten - one.5) / 2 = 0 Miii = 2x + fifteen - 1.five(x - 1.5)2 kN-m These results are only adept for section three (three ≤ x ≤ 4.5m). When ten = 3, the shear force will be, V3 = 6.5 - (three)three = -2.five kN the moment volition be, Miii = (2)3 + fifteen - i.5(3 - 1.5)two = 17.625 kN-thou When ten = 4.5, the shear forcefulness will exist, V3 = 6.5 - (3)four.5 = -7 kN the moment will exist, One thousand3 = (two)4.v + xv - 1.v(4.5 - ane.5)2 = 10.5 kN-m
		Section 4
Section four		The final cutting is at the far right section of the beam. The free-body diagram for the beam will be as shown on the left. Sum of the forces gives,      ΣFy = 0 seven - 5 - three(3) - V4 = 0 Fiveiv = -7 kN Summing the moments well-nigh the cut edge gives,      ΣMcutting = 0 G4 - (7)ten + 5(x - iii) + 3(3)(x - 3) = 0 Yardfour = (seven)x - 14(x-iii) kN-k These results are but good for section 3 (4.5 ≤ x < 6m). When x = 4.5, the shear forcefulness will exist, V4 = -7 kN the moment will exist, Thou4 = (7)4.5 - fourteen(4.5 - three ) = ten.5 kN-g When x = 6, the shear forcefulness will exist, V4 = -seven kN the moment volition be, M4 = (7)6 - 14(6 - 3) = 0 kN-m
		Shear and Moment Diagrams
		At present that the shear and moment is known for each section of the beam, the results can be plotted. The resulting graphics are called the shear diagram and moment diagram. Since the same x was used for all 3 sections, the each equation for each department tin be easily plotted as shown at the left. Since beam and loading are symmetrical, shear force diagram and bending moment diagram can also be drawn by solving first 2 sections only. For shear moment diagram other half volition exist anti-mirror paradigm of outset 2 sections and for bending moment diagram other half will exist mirror image of beginning two sections which can be seen in the effigy shown on the left.

---

Practice Homework and Test problems now available in the 'Eng Statics' mobile app
Includes over 500 problems with consummate detailed solutions.
Available now at the Google Play Store and Apple App Store.

---

7-55 Draw Shear and Moment Diagrams

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

Share this post