Skip to main content

Bending Stresses in Simple Bending

Bending stress or Flexural stress can be define  as the internal resistance per unit area developed by material of the beam against the external bending moment.
According to Theory of bending, the Bending formula is given by,
M/I = f/y = E/R
Where
M = Bending Moment at a section 
I   = Moment of inertia about Neutral axis (N.A.) 
f   = Bending stress in the fiber at distance of y             form the N.A
y  = Distance of the fiber from N.A. 
R = Radius of Curvature 
E = Young's Modulus 
Within the elastic limit, the bending stress (f) is directly proportional to the distance (y) from the neutral axis.
The bending stress f = [M/I].y
The maximum value of bending stress will be the outermost fiber ie,y = ymax
                   fmax = [M/I]. ymax

We can write this equation as, fmax = [M/Z]
Where Z = I/ymax
and Z is called as Elastic Section Modulus

Note:

1.The stress in the outermost section of beam is computed with the help of section modulus (Z).
            fmax = [M/Z]
2.Moment of resistance of the section
            M = Z.fmax 
3.The neutral axis always passes through the centroid of area of a beam's cross-section, so the moment of Inertia about neutral axis is nothing but moment of Inertia about centroidal bending axis.

Examples:

1.A rectangular cross-section beam of length 4 m rests on supports at each end and carries a uniformly distributed load of 10 kN/m. If the stress must not exceed 8 MPa, what will be a suitable depth for the beam if its width is 100 mm?

Solution:

For a simply supported beam with a uniform distributed load w/m over its full length, the maximum bending moment is
M = (wL.L)/8 
w= 10 kN/m and L = 4m
 maximum bending moment M = (10 × 4x4)/8
                 M = 20 kNm.= 20,000,000 N-mm 
The moment of Resistance M = fmax.Z
The maximum bending stress fmax  = 8 MPa
Hence, the required section modulus is:
Z = M/fmax  = 20,000,000/ 8 =25,00000 mm3
For a rectangular cross-section Z = (bd.d)/6 
                    d.d =6Z/b 
The breadth b = 100 mm
 dxd = 6×2500000/100 = 150,000
d = square root of (150,000) = 387 mm
A suitable beam  with a depth of 400 mm may be provided for safe.

2. A cantilever beam of length 2m fails when a load of 2KN is applied at the free end. If the section is 40mmx60mm, find the stress at the failure.

Solution:

Length of beam = 2m or 2000mm
load at failure = 2KN
Beam Section dimensions = 40mm X 60mm
The moment of inertia of beam =
           I = bd3/12
            = (40) (603)/12  = 7.2X105 mm4
The maximum bending moment will occur at the fixed end of cantilever beam
    M = WL
        = (2)(2) = 4KN-m
The maximum fiber distance from N.A.
   ymax = depth /2= 60/2 = 30mm
Substitute all the above in bending equation,
                M /I= f / y = fmax/ ymax
 

Therefore the maximum bending stress at failure is fmax = 166.67N/mm2

3. A rectangular beam 200mm deep and 300mm wide is simply supported over the span of 8m. What uniformly distributed load per metre the beam may carry, if the bending stress is not exceed 120N/mm2.

Solution:

Length of beam = 8m or 8000mm
Section dimensions = 300mm X 200mm
maximum bending stress =fmax  = 120N/mm2.
condition: uniformly distributed load for simply supported beam
The maximum bending moment in a simply supported beam M = wL2/8
                               = w(8x8)/8= 8wX106

Moment of inertia of a rectangular section,
              I = bd3/12
                = (300) (2003) /12  = 2X108 mm4
The maximum fiber distance from N.A.
             ymax = depth /2= 200/2 =100mm
Substitute all the above in bending equation,
                M /I= f / y = fmax/ ymax
(8wX106) /(2X108) = 120 / 100
 The load intensity w = 3X10000 N/m
                                  = 30kN/m

Assignment:

1.A cantilever beam, 50 mm wide by 150 mm high and 6 m long, carries a load that varies uniformly from zero at the free end to 1000 N/m at the wall. (a) Compute the magnitude and location of the maximum flexural stress. 
(b) Determine the type and magnitude of the stress in a fiber 20 mm from the top of the beam at a section 2 m from the free end. 
2. A rectangular beam 300mm deep is simply supported over a span of 4m. Determine the uniformly distributed load per meter which the beam may carry, if the bending stress should not exceed 120N/mm2.Take I=8x106mm4.
3. Calculate the maximum stress induced in a cast iron pipe of external diameter 40mm,of internal diameter 20mm and length 4m when the pipe is supported at its ends and carries a point load of 80N at its centre.
4. A square beam 20mmx20mm in section and 2m long is supported at the ends. The beam fails when a point load of 400N is applied at the centre of the beam. What uniformly distributed load per meter length will break a cantilever of the same material 40mm wide,60mm deep and 3m long?


Labels:

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                
5 comments
Read more

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
3 comments
Read more

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
1 comment
Read more