SHEAR AND BENDING MOMENT DIAGRAMS BASICS AND TUTORIALS

SHEAR AND BENDING MOMENT DIAGRAMS BASIC INFORMATION
A Tutorials on Shear and Bending Moment Diagram? How To Make Shear and Bending Moment Diagram


In order to plot the shear force and bending moment diagrams it is necessary to adopt a sign convention for these responses. A shear force is considered to be positive if it produces a clockwise moment about a point in the free body on which it acts.

A negative shear force produces a counterclockwise moment about the point. The bending moment is taken as positive if it causes compression in the upper fibers of the beam and tension in the lower fiber. In other words, sagging moment is positive and hogging moment is negative.

The construction of these diagrams is explained with an example given in Figure 2.4.

The section at E of the beam is in equilibrium under the action of applied loads and internal forces acting at E as shown in Figure 2.5. There must be an internal vertical force and internal bending moment to maintain equilibrium at Section E.

The vertical force or the moment can be obtained as the algebraic sum of all forces or the algebraic sum of the moment of all forces that lie on either side of Section E.

The shear on a cross-section an infinitesimal distance to the right of pointAisC55 k and, therefore, the shear diagram rises abruptly from 0 to C55 at this point. In the portion AC, since there is no additional load, the shear remainsC55 on any cross-section throughout this interval, and the diagram is a horizontal as shown in Figure 2.4. 

An infinitesimal distance to the left of C the shear is C55, but an infinitesimal distance to the right of this point the 30 k load has caused the shear to be reduced to C25. 

Therefore, at point C there is an abrupt change in the shear force from C55 to C25. In the same manner, the shear force diagram for the portion CD of the beam remains a rectangle. In the portion DE, the shear on any cross-section a distance x from point D is 

               S = 55 − 30 − 4x D 25 − 4x

which indicates that the shear diagram in this portion is a straight line decreasing from an ordinate of C25 at D to C1 at E. 

The remainder of the shear force diagram can easily be verified in the same way. It should be noted that, in effect, a concentrated load is assumed to be applied at a point and, hence, at such a point the ordinate to the shear diagram changes abruptly by an amount equal to the load.

In the portion AC, the bending moment at a cross-section a distance x from point A isM D 55x. Therefore, the bending moment diagram starts at 0 at A and increases along a straight line to an ordinate of C165 k-ft at point C. 

In the portion CD, the bending moment at any point a distance x from C is M D 55.x C 3/ − 30x. Hence, the bending moment diagram in this portion is a straight line increasing from 165 at C to 265 at D. In the portion DE, the bending moment at any point a distance x from D is M D 55.x C 7/ − 30.X C 4/ − 4x2=2. 

Hence, the bending moment diagram in this portion is a curve with an ordinate of 265 at D and 343 at E. In an analogous manner, the remainder of the bending moment diagram can be easily constructed.

Bending moment and shear force diagrams for beamswith simple boundary conditions and subject to some simple loading are given in Figure 2.6.

What is the function of shear keys in the design of retaining walls?

SHEAR KEYS ON RETAINING WALL TUTORIALS
Civil Engineering Tutorials


In determining the external stability of retaining walls, failure modes like bearing failure, sliding and overturning are normally considered in design. In considering the criterion of sliding, the sliding resistance of retaining walls is derived from the base friction between the wall base and the foundation soils.

To increase the sliding resistance of retaining walls, other than providing a large self-weight or a large retained soil mass, shear keys are to be installed at the wall base. The principle of shear keys is as follows:

The main purpose of installation of shear keys is to increase the extra passive resistance developed by the height of shear keys.

However, active pressure developed by shear keys also increases simultaneously. The success of shear keys lies in the fact that the increase of passive pressure exceeds the increase in active pressure, resulting in a net improvement of sliding resistance.

On the other hand, friction between the wall base and the foundation soils is normally about a fraction of the angle of internal resistance (i.e. about 0.8 ) where is the angle of internal friction of foundation soil. When a shear key is installed at the base of the retaining wall, the failure surface is changed from the wall base/soil horizontal plane to a plane within foundation soil.

Therefore, the friction angle mobilized in this case is instead of 0.8 in the previous case and the sliding resistance can be enhanced.