Summary of Shear force and Bending moments

1. The Shear Force at a section may be defined as "the algebraic sum of the total vertical forces on either side of the cross section".

The unit of shear force is : N,or kN

2. The Bending moment at a section may be defined as "the algebraic sum of the moments about that section of all external forces acting to one side of that section".

The unit of bending moment is : N-mm,or kN-m

3. A Shear Force diagram (SFD) is one which shows the variation of the shear force along the length of the beam.

A bending moment diagram (BMD) is one which shows the variation of the bending moment along the length of the beam.

4.Draw the shearing force, and bending moment diagram for the structure, noting the sign conventions.

Sign convention for Shear and Bending moments

To draw SFD or BMD the negative values are plotted to scale below a horizontal reference line and positive values are plotted above the line.

5.Key points in Shear force and Bending moments Diagrams.

a) Bending Moment is maximum at the point where the Shear Force is zero or where it changes its direction from +ve to -ve or vice versa.

b) The point of contraflexure or point of inflexion is the point where the Bending moment changes its direction from +ve to -ve or vice versa. At this point the magnitude of B.M. is zero.

c) The maximum shear force usually occur at the support or at the point under concentrated load.

d) Shear force is constant in unloaded sections, linearly varies in uniformly loaded sections and parabolic over uniformly varying sections.

e)Bending moment varies linearly over unloaded sections, parabolic over uniformly loaded sections and cubic parabolic over uniformly varying sections.

These variations are shown below -

6. The relation among the intensity of load (w), Shear force (V) and Bending moment (M)

a) Rate of change of Shear Force is equal to Intensity of loading

dF/dx = - w(x)

b) The change in the shear force in between any two points is equal to the area under the load diagram.

c) Rate of change of Bending Moment is equal to Shear force.

dM/dx = F(x)

d)The change in bending moment between any two points equals the area under the shear diagram.

Comments

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E) It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E = s/e Modulus of Rigidity (G) It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G = Ƭ / ø Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube PQRS is subjected to a shearing force F. Let Ƭ be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point S moved to S' and point R moved to R' as shown in fig. The shear strain = The angle of distortion ø ø = RR'/ RQ ---(1) Shear strain = Shear stress /Rigidity modulus

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces. Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns. Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method 1. The points of inflection are located at the mid-height of each column above th

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis. Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us consider on

Prof.Kodali Srinivas

Search This Blog