![Elongation of a Bar of circular tapering section
Elongation of a Bar of circular tapering section
due to self weight:
due to self weight:
=Wx*x/(AxE)
(from =PL/AE )
now Wx=1/3* AxX
where Wx=Wt.of the bar
so = X *x/(3E)
so now
L = X *x/(3E)
= /(3E) Xdx= [/3E ] [X2
/2]
= L2
/(6E)
L
0
L
0
x
L
d
A B
X](https://image.slidesharecdn.com/theoryofstructuresce4gdirectandbendingstresses-250709152738-8aa90874/85/theory-of-structures-CE4G-Direct-and-Bending-Stresses-ppt-18-320.jpg)
theory of structures (CE4G) Direct and Bending Stresses.ppt
- 1. Direct and Bending Stresses By V.N.Kundlikar Visit for more Learning Resources
- 2. Mechanics of Solids Mechanics of Solids Syllabus Syllabus:- Part - A 1. Simple Stresses & Strains:- 1. Simple Stresses & Strains:- Introduction, Stress, Strain, Tensile, Compressive & Shear Stresses, Elastic Limit, Hooke’s Law, Poisson’s Ratio, Modulus of Elasticity, Modulus of Rigidity, Bulk Modulus, Bars of Varying Sections, Extension of Tapering Rods, Hoop Stress, Stresses on Oblique Sections.
- 3. 2. Principle Stresses & Strains:- 2. Principle Stresses & Strains:- State of Simple Shear, Relation between Elastic Constants, Compound Stresses, Principle Planes Principle Stresses, Mohr’s Circle of Stress, Principle Strains, Angle of Obliquity of Resultant Stresses, Principle Stresses in beams.
- 4. 3. Torsion:- 3. Torsion:- Torsion of Circular, Solid, Hollow Section Shafts Shear Stress, Angle of Twist, Torsional Moment of Resistance, Power Transmitted by a Shaft, Keys & Couplings, Combined Bending & Torsion, Close Coiled Helical Springs, Principle Stresses in Shafts Subjected to Bending, Torsion & Axial Force.
- 5. Mechanics of Solids Mechanics of Solids Syllabus Syllabus:- Part - B Part - B 1. Bending Moment & Shear Force:- 1. Bending Moment & Shear Force:- Bending Moment, Shear Force in Statically Determinate Beams Subjected to Uniformly Distributed, Concentrated & Varying Loads, Relation Between Bending Moment, Shear force & Rate of Loading.
- 6. 2. Moment of Inertia:- 2. Moment of Inertia:- Concept Of Moment of Inertia, Moment of Inertia of Plane Areas, Polar Moment of Inertia, Radius of Gyration of an Area, Parallel Axis Theorem, Moment of Inertia of Composite Areas, Product of Inertia, Principle Axes & Principle Moment of Inertia.
- 7. 3. Stresses in Beams:- 3. Stresses in Beams:- Theory of Simple Bending, Bending Stresses, Moment of Resistance, Modulus of Section, Built up & Composite Beam Section, Beams of Uniform Strength. 4. Shear stresses in Beams:- 4. Shear stresses in Beams:- Distribution of Shear Stresses in Different Sections.
- 8. 5. Mechanical Properties of Materials:- 5. Mechanical Properties of Materials:- Ductility, Brittleness, Toughness, Malleability, Behaviour of Ferrous & Non-Ferrous metals in Tension & Compression, Shear & Bending tests, Standard Test Pieces, Influence of Various Parameters on Test Results, True & Nominal Stress, Modes of Failure, Characteristic Stress-Strain Curves, Izod, Charpy & Tension Impact Tests, Fatigue, Creep, Corelation between Different Mechanical Properties, Effect of Temperature, Testing Machines & Special Features, Different Types of Extensometers & Compressemeters, Measurement of Strain by Electrical Resistance Strain Gauges.
- 9. AIM OF MECHANICS OF SOLIDS: AIM OF MECHANICS OF SOLIDS: Predicting how geometric and physical properties of structure will influence its behaviour under service conditions.
- 10. • Strength and stiffness of structures is function of size and shape, certain physical properties of material. •Properties of Material:- Properties of Material:- • Elasticity • Plasticity • Ductility • Malleability • Brittleness • Toughness • Hardness
- 11. INTERNAL FORCE:- STRESS INTERNAL FORCE:- STRESS • Axial Compression • Shortens the bar • Crushing • Buckling n m P P P= A • Axial tension •Stretches the bars & tends to pull it apart • Rupture m n =P/A P P
- 12. • Resistance offered by the material per unit cross- sectional area is called STRESS. = P/A Unit of Stress: Pascal = 1 N/m2 kN/m2 , MN/m2 , GN/m2 1 MPa = 1 N/mm2 Permissible stress or allowable stress or working stress = yield stress or ultimate stress /factor of safety.
- 13. • Strain •It is defined as deformation per unit length • it is the ratio of change in length to original length •Tensile strain = increase in length = (+ Ve) () Original length L Compressive strain = decrease in length = (- Ve) () Original length L P L •Strain is dimensionless quantity.
- 14. Strain Strain Stress Stress Stress- Strain Curve for Mild Steel (Ductile Material) Stress- Strain Curve for Mild Steel (Ductile Material) Plastic state Of material Elastic State Of material Yield stress Point E = modulus of elasticity Ultimate stress point Breaking stress point
- 15. Modulus of Elasticity: Modulus of Elasticity: • Stress required to produce a strain of unity. • i.e. the stress under which the bar would be stretched to twice its original length . If the material remains elastic throughout , such excessive strain. • Represents slope of stress-strain line OA. A O stress strain Value of E is same in Tension & Compression. =E E
- 16. A O • Hooke’s Law:- Up to elastic limit, Stress is proportional to strain =E ; where E=Young’s modulus =P/A and = / L P/A = E ( / L) =PL /AE E
- 17. 9 m x 5 m 3m A B P P(9-x)/9 P(x)/9
- 18. Elongation of a Bar of circular tapering section Elongation of a Bar of circular tapering section due to self weight: due to self weight: =Wx*x/(AxE) (from =PL/AE ) now Wx=1/3* AxX where Wx=Wt.of the bar so = X *x/(3E) so now L = X *x/(3E) = /(3E) Xdx= [/3E ] [X2 /2] = L2 /(6E) L 0 L 0 x L d A B X
- 19. L B D P P L+L B-B D-D POISSONS RATIO:- POISSONS RATIO:- = lateral contraction per Unit axial elongation, (with in elastic limit) L(1+) B(1-) D(1-) = (B/B)/(L/L); = (B/B)/() So B = B; New breadth = B -B = B - B =B(1 - ) Sim.,New depth= D(1- )
- 20. for isotropic materials = ¼ for steel = 0.3 Volume of bar before deformation V= L * B*D new length after deformation L1=L + L = L + L = L (1+ ) new breadth B1= B - B = B - B = B(1 - ) new depth D1= D - D = D - D = D(1 - ) new cross-sectional area = A1= B(1- )*D(1- )= A(1- )2 new volume V1= V - V = L(1+ )* A(1- )2 AL(1+ - 2 ) Since is small change in volume = V =V1-V = AL (1-2 ) and unit volume change = V/ V = {AL (1-2 )}/AL V/ V = (1-2 )
- 21. Composite Sections: Composite Sections: • as both the materials deforms axially by same value strain in both materials are same. s = c = s /Es= c /E (= = L /L) _____(1) & (2) •Load is shared between the two materials. Ps+Pc = P i.e. s *As + c *Ac = P ---(3) (unknowns are s, c and L) Concrete Steel bars
- 22. P P/2 P/2 P • Connection should withstand full load P transferred through the pin to the fork . • Pin is primarily in shear which tends to cut it across at section m-n . • Average shear Stress => =P/(2A) (where A is cross sectional area of pin) • Note: Shearing conditions are not as simple as that for direct stresses. Direct Shear:-- Direct Shear:-- Pin Pin m n Fork For more detail contact us