Skip to main content

Shear Force and Bending Moment Diagrams in Cantilever Beams

Q.1 Draw the shear force and bending moment diagrams of a cantilever beam carrying a point load W at its free end as shown in figure. 

Solution:

For cantilevered beams we need not find support reactions, if we considering the free-end of the beam as the initial starting point of the analysis.

Let us consider one section XX at a distance x from free end i.e. from end B. 

Now we will have two portions of the beam AB i.e. left portion and right portion. 

Let us deal with the right portion and draw the free body diagram.

(you can also go with left portion of the beam, you required to find the support reactions ) 

The section is valid in between B and A, becouse of there is no load in between the points.

Let us assume that Vx is shear force and Mx is the bending moment at section xx. 
Shear force at section xx will be equal to the resultant force acting to the right portion of the section. 
Vx = + W ---(1)
We will recall here the sign conventions for shear force and bending moment and we can conclude here that resultant force acting to the right portion of the section will be W and it will be positive.
When x=0 ,the shear force at free end B is 
VB = +W
When x=L ,the shear force at fixed end A is
VA = +W 

Bending moment at section XX will be written as
Mx = - W. x   -- (2)
Above equation indicates that bending moment will follow here the linear equation.
Again if we will recall sign conventions for shear force and bending moment, we will conclude here that bending moment at section xx will be negative.
Bending moment will be directionally proportional with the distance x and we can secure here the value of bending moment at point A an at point B as mentioned here.
Bending moment at free end i.e. at B, value of distance x = 0
MB= 0
Bending moment at fixed end i.e. at A, value of distance x = L
MA= - W.L
After calculating shear force and bending moments values at critical points the SFD and BMD are drawn below.
(The negative values are plotted to scale below a horizontal reference line and positive values are plotted above the line.)
Shear Force and Bending Moment Diagrams

Q.2 Draw the shear force and bending moment diagrams of a cantilever beam carrying  point loads as shown in figure. 

Solution:

For cantilevered beams we need not find support reactions, if we considering the free-end of the beam as the initial starting point of the analysis.
Segment AB
Let us consider one section XX in between A and B,at a distance x from free end A.
The limits of x is zero to 1 meter.
The S.F. equation Vx = +10 KN
The B.M. equation Mx = - 10.X
section XX in between A and B
When x = 0,
Va = 10 KN, and Ma = 0
When x = 1m,
Vb = 10 KN. and Mb = -10 KN-m
Segment BC
Now we will consider another section XX in between B and C,at a distance x from free end A, and consider right side portion.
The limits of x is 1 meter to 2 meters.
The S.F. equation 
Vx = (+10 + 20 )KN = + 30 KN
The B.M. equation Mx = - 10.X - 20 (x- 1) 
section XX in between B and C
When x= 1m,
Vb = 30 KN, and Ma = - 10
When x = 2m,
Vb = 30 KN. and Mb = - 40 KN-m
Segment CD
Now we will consider anther section XX in between C and D,at a distance x from free end A.
The limits of x is 2 meter to 3 meters.
The S.F. equation 
Vx = (+10 + 20 +20 )KN = + 50 KN
The B.M equations 
Mx = - 10.X - 20 (x- 1) - 20 (x- 2)
section XX in between C and D
When x= 2m,
Vb = +50 KN, and Ma = - 40 KN-m
When x = 3m,
Vb = +50 KN. and Mb = - 90 KN-m
Plot the SFD and BMD by using the above values.
Loading Diagram
Shear Force Diagram

Bending Moment Diagram


Q.3 Draw the shear force and bending moment diagrams of a cantilever beam carrying uniformly distributed load w (N/m) as shown in figure.
Solution :

For cantilevered beams we need not find support reactions, if we considering the free-end of the beam as the initial starting point of the analysis.
Let us consider one section XX in between A and B,at a distance x from free end B and consider right side portion.

The limits of x is zero to L.
The S.F. equation at the section,Vx = +wx
The B.M. equation at the section, Mx = - wx. {x/2}

The above B.M. equation is represented parabola equation, 

When x= 0,
The shear force at B is Vb = 0, 
and The bending moment Mb = 0 
When x = L,
The shear force at B is Va = +wL
and The bending moment Ma = - w.L. L/2
Plot the SFD and BMD by using the above values.

Assignment :
Q1.1. Draw the shear force and bending moment diagrams of a cantilever beam carrying  point loads as shown in figure. 
Q1.2. Draw the shear force and bending moment diagrams of a cantilever beam carrying  a point load as shown in figure.
Q1.1. Draw the shear force and bending moment diagrams of a cantilever beam carrying  uniformly distributed load as shown in figure.

Comments

Post a Comment

Popular posts from this blog

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