Understanding Internal Loadings in Engineering Mechanics and Beams
Learn how to determine internal loadings in beams and mechanical members by investigating external loads and reactions. Discover the role of normal force, shear force, and bending moment, along with sign conventions and the effects of bending and shear in beams.
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STATICS (ENGINEERING MECHANICS-I) Beams-Internal Effects June 7, 2025 1
Beams-Internal Effects The design of any structural or mechanical member requires an investigation of both the external loads and reactions acting on the member and the loading acting within the member in order to be sure the material can resist this loading. How to determine the Internal Loadings: The internal loadings can be determined using the method of sections. The idea is to cut an imaginary section through the member so that the internal loadings (of interest) at the section become external on the free body diagram of the section. Use equations of equilibrium and find the required internal forces. 6/7/2025 2
Internal forces If the member is subjected to a coplanar system of forces, only. N (Normal force), V (shear force) , and M (bending moment) act at the section. Note: N will induce when applied loading is oblique. V M N N M V N: Normal force; V: Shear force; M: Bending moment. 6/7/2025 4
Sign Convention M V N N V M The directions shown for N, V and M are in a positive sense. The easy way to remember this sign convention is to isolate a small segment of the member and note that: Positive normal force tends to elongate the segment. N N Positive shear tends to rotate the segment clockwise. V V Positive bending moment tends to bend the segment concave upward, so as to hold water. M M 6/7/2025 6
Effect of Bending and Shear 6/7/2025 7
Shear Force and Bending Moment in a Beam To determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads. Determine reactions at supports by treating whole beam as free-body. Cut beam at C and draw free-body diagrams for AC and CB. By definition, positive sense for internal force-couple systems are as shown. From equilibrium considerations, determine M and V or M and V . 5- 8
Shear-force and Bending-moment diagrams The variation of shear-force and bending moments plotted against distance along the beam give the shear-force and bending moment diagrams for the beam. 6/7/2025 9
Shear Force and Bending Moment Diagrams Variation of shear and bending moment along beam may be plotted. Determine reactions at supports. Cut beam at C and consider member AC, 2 M P V + = = + 2 Px Cut beam at E and consider member EB, 2 M P V = ( ) 2 = + P L x + _ For a beam subjected to concentrated loads, shear force is constant between loading points and bending moment varies linearly. + + 5- 10
NOTE The part of the beam which involves the smaller number of forces, either to the right or to the left of the arbitrary section, usually yields the simpler solution. Avoid using a transverse section that coincides with the location of a concentrated load or couple, as such this position represents a point of discontinuity in the variation of shear or bending moment. 6/7/2025 11
Problem-1 Determine the shear and moment values in the simple beam at three sections: 1.5 m, 2.5 m and 4 m from the left support. 4 kN 2 kN/m 2.8 kN.m B A 1 m 1 m 1 m 1 m 1 m 6/7/2025 12
Solution Reactions: 4 kN 2 kN/m 2.8 kN.m A x A B 1 m 1 m 1 m 1 m 1 m y A B y + = = 0 0 F A x x + = + = + = 0 2 1 4 0 6 F A B A B y y y B y y + = ) 1 5 . 2 + = ( ) 0 5 4 4 2 ( 8 . 2 0 CCW M A y = = = . 3 64 kN 6 . 3 64 . 2 36 kN B A y y 6/7/2025 13
SF and BM calculations 5 . 1 = m x = 5 . 2 m x 2 kN/m 2.8 kN.m V 2.8 kN.m V A M A o o 1.5 2.5 M 2 m . 2 36 kN . 2 36 kN + = 0 Fy + = 0 Fy = . 2 V 36 = 5 . 2 ( 2 ) 2 0 V = . 2 V 36 = 0 V . 1 36 + kN M . 2 36 kN = ( ) 0 CCW O + = ( ) 0 CCW M 5 . 2 + + . 2 36 8 . 2 M O 5 . 1 + = . 2 36 8 . 2 0 M 5 . 0 5 . 0 = 2 0 = kN.m 74 . 0 M 2 = . 2 kN.m 85 M 6/7/2025 14
SF and BM calculations Section just after the point load Section just before the point load 4kN 2 kN/m 2.8 kN.m 2 kN/m V 2.8 kN.m M V M o A 1 m 1 m 1 m o 4m . 2 36 kN 1 m 1 m 1 m 4 m . 2 36 kN + = + = 0 0 Fy Fy = = . 2 V 36 = 2 1 0 . 2 V 36 = + 2 1 4 0 V V . 3 M . 0 + 36 M kN 64 kN = = ( ) 0 + ( ) 0 + CCW CCW O O + ) 1 4 ( = + ) 1 4 ( + = . 2 36 4 8 . 2 2 ( ) 5 . 2 0 . 2 36 4 8 . 2 2 ( ) 5 . 2 4 0 0 M M = = 3.64 kN.m kN.m 64 . 3 M M BM is the same under the point load but SF changes from 0.36 kN to -3.64 kN. 6/7/2025 15
