Relation between load shear force and bending moment of beams

Relation between load shear force and bending moment

1. BEAMS Relation between Load S.F and B.M Mrs. Venkata Sushma Chinta

2. Free body diagram of the beam         - *(1)- 1.5 *2 * (2 +1 ) =0 =- ( 4+ 1.5 *2 * 3) = - 13 kN-m   Q4. Calculate the support reactions for the beam as shown in the figure

3. Free body diagram of the beam         *(1)- ( ) * (1+ ) - =0 = ( 350 + 875 * ) = 904.2 kN   Magnitude of UVL= Area of loading Diag = = = 875 N         => => 320.8 kN   Q5. Calculate the support reactions for the beam as shown in the figure

4. Finding the reactions is usually the first step in the analysis of a beam. Once the reactions are known, the shear forces and bending moments can be found. If a beam is supported in a statically determinate manner, all reactions can be found from free-body diagrams and equations of equilibrium. Internal stress resultants: Shear forces and bending moments are the resultants of stresses distributed over the cross section. Therefore, these quantities are known collectively as stress resultants. The stress resultants in statically determinate beams can be calculated from equations of equilibrium. When designing a beam, we usually need to know how the shear forces and bending moments vary throughout the length of the beam. Of special importance are the maximum and minimum values of these quantities. Information of this kind is usually provided by graphs in which the shear force and bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as the abscissa. Such graphs are called shear-force and bending-moment diagrams.

5. BEAMS SHEAR FORCE AND BENDING MOMENTS

6. Shear forces and bending moments , like axial forces in bars and internal torques in shafts, are the resultants of stresses distributed over the cross section. Therefore, these quantities are known collectively as stress resultants. The stress resultants in statically determinate beams can be calculated from equations of equilibrium.

7. When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam. To determine these stresses and strains, we first must find the internal forces and internal couples that act on cross sections of the beam.

8. Relation between loads, shear force and bending moments: The relationships between loads, shear forces, and bending moments in beams are quite useful when investigating the shear forces and bending moments throughout the entire length of a beam, and they are especially helpful when constructing shear-force and bending-moment diagrams. Element of a beam used in deriving the relationships between loads, shear forces, and bending moments

9. q Distributed load: From this equation we see that the rate of change of the shear force at any point on the axis of the beam is equal to the negative of the intensity of the distributed load at that same point.

10. Type of distributed Load Shear force V Variation of Shear force No distributed load on the segment of beam (q=0) V = C The shear force remains constant in that part of the beam if the distributed load is uniform along part of the beam (q= C) V = C. The shear force varies linearly in that part of the beam If linearly varying load acting on segment of beam (q= ) V = The shear force varies quadratic in that part of the beam Type of distributed Load Variation of Shear force No distributed load on the segment of beam (q=0) The shear force remains constant in that part of the beam if the distributed load is uniform along part of the beam (q= C) The shear force varies linearly in that part of the beam The shear force varies quadratic in that part of the beam   V      

11. Distributed load: Discarding products of differentials (because they are negligible compared to the other terms), we obtain the following relationship: This equation shows that the rate of change of the bending moment at any point on the axis of a beam is equal to the shear force at that same point.

12. Type of distributed Load Shear force V Variation of Shear force Bending Moment M Variation of bending moment No distributed load on the segment of beam (q=0) = C remains constant M = C. The Bending moment varies linearly if the distributed load is uniform along part of the beam (q= q) = q. shear force varies linearly M = The Bending moment varies quadratically If linearly varying load acting on segment of beam (q= ) = The shear force varies quadratic in that part of the beam M = The Bending moment varies cubically Type of distributed Load Variation of Shear force Variation of bending moment No distributed load on the segment of beam (q=0) remains constant The Bending moment varies linearly if the distributed load is uniform along part of the beam (q= q) shear force varies linearly The Bending moment varies quadratically The shear force varies quadratic in that part of the beam The Bending moment varies cubically

13. Concentrated Loads: Since the length dx of the element is infinitesimally small, we see from this equation that the increment M 1 in the bending moment is also infinitesimally small. Thus, the bending moment does not change as we pass through the point of application of a concentrated load. This result means that an abrupt change in the shear force occurs at any point where a concentrated load acts. As we pass from left to right through the point of load application, the shear force decreases by an amount equal to the magnitude of the downward load P.

14. Load in the form of couple: From equilibrium of the element in the vertical direction we obtain V 1 = 0, which shows that the shear force does not change at the point of application of a couple. Equilibrium of moments about the left-hand side of the element gives This equation shows that the bending moment decreases by M as we move from left to right through the point of load application. Thus, the bending moment changes abruptly at the point of application of a couple.