Analysis of Beams and FramesTheory of Structure - I

Lecture OutlinesShear and Moment Diagrams for Beams Shear and Moment Diagrams for a Frames Moment Diagrams Constructed by the Method of Superposition Deflected Curves

Shear and Moment Diagrams for a Beam

Dividing by Dx and taking the limit as Dx 0, these equation becomeSlope of Moment Diagram = ShearSlope of Shear Diagram = -Intensity of Distributed Load

Equations (4-1) and (4-2) can be integrated from one point to another between concentrated forces or couples, in which caseChange in Shear = -Area under Distributed Loading DiagramChange in Moment = Area under Shear Diagram

Thus, when F acts downward on the beam, DV is negative so that the shear diagram shows a jump downward. Likewise, if F acts upward, the jump (DV) is upward.In this case, if an external couple moment M is applied clockwise, DM is positive, so that themoment diagram jumps upward, and when M acts counterclockwise, the jump (DM) must bedownward.

Example 4-7

Draw the shear and moment diagrams for the beam shown in the figure.

SOLUTIONx = 5.20 mM = 104 kNm= 5.20 m

Example 4-8

Draw the shear and moment diagrams for each of the beam shown in the figure.

SOLUTION

Example 4-9

Draw the shear and moment diagrams for the beam shown in the figure.

SOLUTION

Example 4-10

Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A is fix C is roller and B is pin connections.

SOLUTION

Example 4-11

Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A and C are rollers and B and D are pin connections.

SOLUTION

Shear and Moment Diagrams for a FrameP1P2AyCxCy

Example 4-12

Draw the shear and moment diagrams for the frame shown . Assume A, C and D are pinned and B is a fixed joint.

Find the Reaction

Member AB

Member BC

Member CD

Bending moment diagram of frame

Example 4-13

Draw the moment diagram for the frame shown . Assume A is pin, C is a roller, and B is a fixed joint.

-Bxcos 36.87 + Bysin 36.87 + 0 = 0-----(1)-Bxsin 36.87 - Bycos 36.87 + 2.5 = 0-----(2)

Moment Diagrams Constructed by the Method of SuperpositionMost loading on beams in structural analysis will be a combination of the loadings shown in the figure below:

Example 4-14

Draw the moment diagrams for the beam shown at the top of the figure below using the method of superposition. Consider the beam to be cantilevered from the support at B.

SOLUTION

Deflected Curve