Skip to main content

Bending stresses in Beams

When a beam is subjected to external loads, shear forces and bending moments develop in the beam. Therefore, a beam must resist these external shear forces and bending moments. The beam itself must develop internal resistance to resist shear forces and bending moments.
The stresses caused by the bending moments are called bending stresses. For beam design purposes, it is very important to calculate the shear stresses and bending stresses at various locations of a beam. 
The bending stress varies from zero at the neutral axis to a maximum at the tensile and compressive side of the beam.

Procedures for determining bending stresses 

Stress at a Given Point: 

1.Use the method of sections to determine the bending moment M at the cross section containing the given point. 
2. Determine the location of the neutral axis.
3. Compute the moment of inertia I of the cross- sectional area about the neutral axis.
4. Determine the y-coordinate of the given point. Note that y is positive if the point lies above the neutral axis and negative if it lies below the neutral axis.
5.Compute the bending stress from σ = -My / I. If correct sign are used for M and y, the stress will also have the correct sign (tension positive compression negative).

Bending Stress in
Symmetrical Cross Sections 

Flexural Stress varies directly linearly with distance from the neutral axis. Thus for a symmetrical sections the compressive and tensile stresses will be the same. This will be desirable if the material is both equally strong in tension and compression. Symmetrical bending stresses arises in beams which have the bending axis (N.A) coincide with either of symmetrical axis. Singly or doubly symmetrical cross sections,  are shown below.


If the neutral axis is an axis of symmetrical of the cross section, the maximum tensile and compression bending stresses are equal in magnitude and occur at the section of the largest bending moment. 
The following procedure is recommended for determining the maximum bending stress in a prismatic beam: 
1.Draw the bending moment diagram and Identify the maximum bending moment Mmax (disregard the sign) 
2.Compute the moment of inertia I of the cross- sectional area about the neutral axis. 
3. Calculate the maximum bending stress from
 σ max = [Mmax]ymax / I, 
where ymax is the distance from the neutral axis to the top or bottom of the cross section.

Unsymmetrical Cross Sections:

Symmetrical sections will be desirable if the material is both equally strong in tension and compression. However, some materials are strong in compression than in tension. It is therefore desirable to use a beam with unsymmetrical cross section giving more area in the compression part making the stronger fiber located at a greater distance from the neutral axis than the weaker fiber. Unsymmetrical bending stress arises in beams which have a bending axis is not symmetrical about cross sections. Some of these sections are shown below.

If the neutral axis is not an axis of symmetry of the cross section, the maximum tensile and compressive bending stresses may occur at different sections. 
1. Draw the bending moment diagram. Identify the largest positive and negative bending moments. 
2. Determine the location of the neutral axis and record the distances y top and y bottom from the neutral axis to the top and bottom of the cross section. 
3. Compute the moment of inertia I of the cross section about the neutral axis. 
4. Calculate the bending stresses at the top and bottom of the cross section where the largest positive bending moment occurs from σ = -My / I.  At the top of the cross section, where y = ytop,we obtain σtop = -Mytop/ I. 
At the bottom of the cross section, we have y = - ybot, so that σbot = Mybot/ I. 
5.Repeat the calculations for the cross section that carries the largest negative bending moment. 
6.Inspect the four stresses thus computed to determine the largest tensile (positive) and compressive (negative) bending stresses in the beam.

Comments

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           ...

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 co...

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 e...