Module 4 flexural stresses- theory of bending

INTRODUCTION TO SOLID MECHANICS (UGCE) 4TH SEM

1. Mr. Akash Assistant Professor Department of Civil Engineering DCE Darbhanga Module- 4 Flexural Stresses-Theory of simple bending Introduction to Solid Mechanics Course Code- 101205

2. Syllabus Flexural Stresses-Theory of simple bending Assumptions – Derivation of bending equation: M/I = f/y = E/R - Neutral axis – Determination of bending stresses – Section modulus of rectangular and circular sections (Solid and Hollow), I,T, Angle and Channel sections – Design of simple beam sections.

3. Behaviour of Beam If we take any section of a beam, as in figure (a ), and consider the free body on either side, we find that there must be three internal stress resultants at the section in general to maintain the equilibrium of the free body. These internal stress resultants are equal and opposite to the external effects of loads on the section, i.e. the bending moment, shear force, and axial force as in figure (b) Under the effect of the lateral loads, the beam bends and the longitudinal axis takes a curved shape. The beam produces an internal resisting couple to balance the BM due to the applied loads. This couple is generated by the normal compressive and tensile forces in the section, as shown in Figure (d). This couple is equal and opposite to the external BM . To generate a force to balance the SF due to applied loads, the section has tangential stresses whose resultant is equal and opposite to the SF (Figure (e)

4. Bending(Flexure) Stresses When a beam is subjected to a loading system or by a force couple acting on a plane passing through the axis, then the beam deforms. In simple terms, this axial deformation is called as bending of a beam. Due to the shear force and bending moment, the beam undergoes deformation. These normal stress due to bending are called flexure stresses. Type of Bending Stress 1. Simple Bending Stress Bending will be called as simple bending when it occurs because of beam self-load and external load. This type of bending is also known as ordinary bending and in this type of bending results both shear stress and normal stress in the beam.

5. Pure Bending Stress Bending will be called as pure bending when it occurs solely because of coupling on its end. In that case there is no chance of shear stress in the beam. But, the stress that will propagate in the beam as a result will be known as normal stress. Normal stress because it not causing any damages to beam. As shown below in the picture, pure Bending stresses are those that results because of beam self load only .

6. Theory of Pure Bending: Bernoulli's Equation Consider a beam bent by pure couples so that the BM is constant and the SF is zero. The beam bends in such cases in the form of an arc of constant radius. The beam is prismatic, i.e., of constant cross section and is symmetrical about a vertical plane through its centroid. In deriving the theory of bending, the following assumptions are made Sections which are plane before bending remain plane after bending. sections remain plane, there must be one layer which is neither stretched nor shortened. The plane of loading must contain a principal axis of the beam cross section. The plane containing the axis of symmetry should also contain the loads and reactive forces. The material is homogeneous and isotropic and obeys Hooke’s law. The modulus of elasticity of the materials is the same in tension and compression. Stress diagram is linear and has the same shape as the strain diagram. The beam is initially straight and has constant cross section, i.e., a prismatic beam. Longitudinal axis of beam is straight. Each layer of the material is free to expand or contract under stress. This is necessary to ensure that we are able to calculate the stress at any layer using the elastic properties. The layers are not affected by the presence of adjoining layers, and are free to deform. Beam is initially straight, self weight of beam is not considered and Poisson effect in beam is neglected.

7. I n bending, one surface of the beam is subjected to tension and the opposite surface to compression there must be a region within the beam cross-section at which the stress changes sign, i.e. where bending stress should be zero. The layer of zero stress due to bending is called neutral layer and the trace of neutral layer in the cross-section is called neutral axis. Due to pure bending, beams sag or hog depending upon the nature of bending moment. It can be easily observed that when beams sag, fibres in the bottom side get stretched while fibres on the top side are compressed. In other words, the material of the beam is subjected to tensile stresses in the bottom side and to compressive stresses in the upper side. In case of hogging the nature of bending stress is exactly opposite, i.e ., tension at top and compression at bottom. Neutral Axis

8. Bending Equation Based upon these assumptions, the bending equation can be derived as follows . Let us consider a beam initially unstressed as shown in figure. Now the beam is subjected to a constant bending moment (i.e. ‘Zero Shearing Force') along its length as would be obtained by applying equal couples at each end. The beam will bend to the radius R as shown in Fig below. The layers above n-n are compressed and those below n-n are stretched in the case shown of positive BM . If the BM is negative, the upper layers will be in tension and the bottom layers will be in compression, with the beam bending convex upwards. R is the radius to which the neutral layer bends. Considering the deformation of a layer y above n-n, we have dx= R d θ where d θ is the angle subtended by the length dx at the centre of curvature, and m-m = (R - y)d θ . The length of the layer m-m before bending is dx = R d θ . Change in length = R d θ - (R - y)d θ =y dθ

9. and Since R is a constant, strain at any layer is proportional to its distance y from the neutral layer. Since E is constant and is the same in tension and compression, stress σ at layer m-m = strain x E . or, …………………(1) The stress diagram is thus linear following the variation in strain. This diagram is obtained by multiplying the strain diagram by the constant E.

10. Considering an elementary strip dy of width b of the cross section, the force acting on the elementary area is given by Or, from equation (1) This can be summed up for the entire cross section and since C = T, such summation is equal to zero. If y 1 and y 2 , are the extreme fibre depths above and below the NA, then Moment of dF about the neutral axis would be , dM = dF y

11. Total Moment for the whole cross-section is therefore equal to is the property of the material and is called as a second moment of area of the cross-section and is denoted by a symbol “I” Therefore, or, Since, Or, This equation is known as the Bending Equation or flexural formulae. The above proof has involved the assumption of pure bending without any shear force being present. Therefore this termed as the pure bending equation. This equation gives distribution of stresses which are normal to cross-section i.e. in x-direction.

12. Locating Neutral Axis Consider an elemental area δ A at a distance y from neutral axis If ‘ σ ’ is the stress on it, force on it F= σ δ A But σ= Ey /R , from eqn..(1) ∴ Force on the element, Hence total horizontal force on the beam Since there is no other horizontal force, equilibrium condition of horizontal forces gives As is not zero . …………………..(2) If A is total area of cross-section, from eqn. (2), we get is moment of area about neutral axis is distance of centroid of the area from the neutral axis. Hence means the neutral axis coincides with the centroid of the cross-section

13. Section Modulus Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. Section Modulus is moment of inertia of the section divided by the distance to the extreme fibre, either to the top or to the bottom, from the neutral axis. Section modulus is usually denoted by the symbol Z . The unit of section modulus Z is From Bending equation i.e. Hence bending stress is maximum, when y is maximum. It means maximum stress occurs in the extreme fibres. If is distance of extreme fibres from neutral fibre, then In a design is restricted to the permissible stress in the material. If is permissible stress , then from above equation

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16. 5 6. I – Section

17. Example : 1. A timber beam has to carry a load of 2 kN /m over a span of 3 m. The permissible stresses are 12 MPa in compression and 8 MPa in tension. Design the section if the width is half of the depth . Solution :

18. Example : 2. A steel beam has a span of 16 m and carries a load of 14 kN /m. Design a suitable I-section if the permissible stress is 150 MPa . Solution :

19. Example : 3. S hows the cross-section of a cantilever beam of 2.5 m span. Material used is steel for which maximum permissible stress is 150 . What is the maximum uniformly distributed load this beam can carry ? Solution : Since it is a symmetric section, centroid is at mid depth. I = MI of 3 rectangles about centroid

20. Load Carrying Capacity Given a beam section and the span on which it is required to support loads, we can find the load that the section can carry. This is generally done for UD load. More appropriate is the moment carrying capacity called moment of resistance of a section. This is independent of the span and the loads. Given a section and the maximum permissible stresses in tension and compression, the moment carrying capacity can be calculated. This can be compared with the maximum moment in a span and the load condition. The examples to illustrate both the load carrying capacity and moment of resistance are given later Question : A rectangular section, 200 mm x 400 mm, spans a distance of 2 m. Find what UD load can the beam section carry on this span. Permissible stresses are 7 in tension and 15 in compression. Solution:

21. Example: 1 A circular log of wood is used as a beam. If the diameter of the log is 200 mm, find the moment of resistance of the section. Permissible stresses are 10 in tension and 18 in compression. Solution: Example: 2. Design a rectangular section to carry a load of 32 kN /m over its whole span of 3 m, including the self- weight. Permissible stresses are 8 in tension and 15 in compression. Solution:

22. Example: 3. The section shown in Fig. 5.34 is a reinforced concrete beam section. Find the moment capacity of the section if the permissible stress in concrete is 5 in compression and 140 in steel in tension . The modular ratio is 18 . Solution:

23. Example: 4. Find the moment resistance of the section shown in Figure. Permissible stresses are 100 MPa in tension and compression . What UD load can it carry over a span of 8 m? Solution :

24. Example: 5 . An I-section girder, 200 mm wide by 300 mm deep, with flange and web of thickness 20 mm is used as a simply supported beam over a span of 7 m. The girder carries a distributed load of 5 kN /m and a concentrated load of 20 kN at mid-span. Determine: (a) the second moment of area of the cross-section of the girder, ( b) the maximum stress set-up . Solution :

25. Example: 6. A uniform T-section beam is 100 mm wide and 150 mm deep with a flange thickness of 25 mm and a web thickness of 12 mm. If the limiting bending stresses for the material of the beam are 80 in compression and 160 in tension, find the maximum u.d.l . that the beam can carry over a simply supported span of 5 m. Solution :

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