Beams SFD and BMD
[Pages:21]Beams ? SFD and BMD
Shear and Moment Relationships
w dV dx
Slope of the shear diagram = - Value of applied loading
V dM dx
Slope of the moment curve = Shear Force
Both equations not applicable at the point of loading because of discontinuity produced by the abrupt change in shear.
ME101 - Division III
Kaustubh Dasgupta
1
Beams ? SFD and BMD
w dV dx
V dM dx
Degree of V in x is one higher than that of w Degree of M in x is one higher than that of V
Degree of M in x is two higher than that of w
Combining the two equations
d 2M dx 2
w
M :: obtained by integrating this equation twice
Method is usable only if w is a continuous function of x (other cases not part of this course)
ME101 - Division III
Kaustubh Dasgupta
2
Beams ? SFD and BMD
Shear and Moment Relationships
Expressing V in terms of w by integrating w dV dx
V
x
dV wdx
V0
x0
OR V = V0 + (the negative of the area under the loading curve from x0 to x)
V0 is the shear force at x0 and V is the shear force at x
Expressing M in terms of V by integrating V dM
dx
M
x
dM Vdx
M0
x0
OR M = M0 + (area under the shear diagram from x0 to x)
M0 is the BM at x0 and M is the BM at x
ME101 - Division III
Kaustubh Dasgupta
3
Beams ? SFD and BMD
V = V0 + (negative of area under the loading curve from x0 to x)
M = M0 + (area under the shear diagram from x0 to x)
If there is no externally applied moment M0 at x0 = 0, total moment at any section equals the area under the shear diagram up to that section
When V passes through zero and is a continuous function of x with dV/dx 0 (i.e., nonzero loading)
dM 0 dx BM will be a maximum or minimum at this point
Critical values of BM also occur when SF crosses the zero axis discontinuously (e.g., Beams under concentrated loads)
ME101 - Division III
Kaustubh Dasgupta
4
Beams ? SFD and BMD: Example (1)
? Draw the SFD and BMD.
ME101 - Division III
? Determine reactions at supports.
? Cut beam at C and consider member AC,
V P 2 M Px 2
? Cut beam at E and consider member EB,
V P 2 M PL x 2
? For a beam subjected to
Maximum BM occurs where Shear changes the direction
concentrated loads, shear is constant between loading points and moment varies
linearly
Kaustubh Dasgupta
5
Beams ? SFD and BMD: Example (2)
Draw the shear and bending moment diagrams for the beam and loading shown. Solution: Draw FBD and find out the support reactions using equilibrium equations
ME101 - Division III
Kaustubh Dasgupta
6
SFD and BMD: Example (2)
Use equilibrium conditions at all sections to
get the unknown SF and BM
Fy 0 :
20 kN V1 0
V1 20 kN
M1 0 : 20 kN0 m M1 0 M1 0
V2 = -20 kN; M2 = -20x kNm
V3 = +26 kN; M3 = -20x+46?0 = -20x MB= -50 kNm V4 = +26 kN; M4 = -20x+46?(x-2.5) MC= +28 kNm V5 = -14 kN; M5 = -20x+46?(x-2.5)-40?0 MC= +28 kNm
V6 = -14 kN; M6 = -20x+46?(x-2.5)-40?(x-5.5) MD= 0 kNm
ME101 - Division III
Kaustubh Dasgupta
7
Beams ? SFD and BMD: Example (3)
Draw the SFD and BMD for the beam acted upon by a clockwise couple at mid point
C
Solution: Draw FBD of the beam and
l
Calculate the support reactions
V
C C l
Draw the SFD and the BMD starting
C l
From any one end
M
ME101 - Division III
Kaustubh Dasgupta
C 2
C 2
8
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