Beams - SFD and BMD - Lecture 11

Oct 25, 2026

Uploaded by Marglet

Lecture notes on beam analysis using shear force and bending moment diagrams. Covers shear–moment relationships, including how the slope of the shear diagram relates to applied loading and how integrating shear or loading functions yields expressions for shear force V and bending moment M. Discusses conditions for validity when the load intensity is continuous, treatment of discontinuities at points of applied loading or concentrated loads, and how maximum or minimum bending moment occurs when shear changes sign. Includes step-by-step guidance for drawing SFD and BMD and solving example problems using equilibrium.

LanguageEnglish

CategoryTechnology

FormatPDF

Pages21

Reading Time53 min

AccessFree to download

Shear and Moment Relationships

Relations among w, V, and M

Validity conditions and discontinuities

Drawing SFD and BMD for beams

Worked examples

Beams – SFD and BMD Shear and Moment Relationships w 􀀠 􀀐 ~~ ~~ddVx Slope of the shear diagram = -Value of applied loading dM V 􀀠 ~~ ~~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 􀀠 􀀐 ~~ ~~ddVx dM V 􀀠 ~~ ~~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 2 Mdx 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 􀀠 􀀐 ~~ ~~ddVx 􀂳VV 0 dV 􀀠 􀀐 􀂳xx 0wdx 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 􀀠 dMdx 􀂳MM 0dM 􀀠 􀂳xx 0Vdx OR M = M0 + f( area undrom x 0toerx)the shear diagram M0 is the BM at x0 and M is the BM at x ME101-Division III Kaustubh Dasgupta 3 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 Maximum BM occurs where Shear changes the direction • Draw the SFD and BMD. • Determine reactions at supports. • Cut beam at C and consider member AC, V 􀀠 􀀎 P2 M 􀀠 􀀎 Px2 • Cut beam at E and consider member EB, V 􀀠 􀀐 P2 M 􀀠 􀀎 P􀀋L􀀐 x􀀌2 • For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly ME101-Division III Kaustubh Dasgupta 5 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 kN􀀌􀀋0 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 Draw the SFD and BMD for the beam acted upon by a clockwise couple at mid point Solution: Draw FBD of the beam and Calculate the support reactions Draw the SFD and the BMD starting From any one end C l V C l C C 􀀐 l 2 ME101-Division III Kaustubh Dasgupta 8 Beams – SFD and BMD: Example (3) Draw the SFD and BMD for the beam Solution: Draw FBD of the beam and Calculate the support reactions ∑MA = 0 􀃆 RA = 60 N ∑MB = 0 􀃆 RB = 60 N Draw the SFD and the BMD starting from any one end | | 120 Nm | | | --- | --- | --- | | | 60 N | | | | | | | | | | | | | | | | -120 Nm | | 60 N V -60 N M ME101-Division III Kaustubh Dasgupta 9 Draw the SFD and BMD for the beam Solution: Draw FBD of the entire beam and calculate support reactions using equilibrium equations Reactions at supports: RA 􀀠 RB 􀀠 wL2 Develop the relations between loading, shear force, and bending moment and plot the SFD and BMD ME101-Division III Kaustubh Dasgupta 10

FAQs

How are the slope of the shear diagram and applied loading related?

The slope of the shear diagram equals the negative value of the applied loading. This establishes the link between w(x) and the shear-force function V(x).

How can V and M be expressed using integration?

V can be obtained by integrating the loading function (or from the area under the loading curve, with sign). M can be obtained by integrating shear (or from the area under the shear diagram), using integration constants defined at x0.

When do bending moment maxima or minima occur?

Bending moment reaches a maximum or minimum where V passes through zero and V is continuous with nonzero dV/dx (i.e., nonzero loading). Critical values can also occur at discontinuous crossings of the shear axis for cases like concentrated loads.

Oct 25, 2026

Uploaded by Marglet