Mmax occurs at places where V = 0 or V changes sign.This occurs twice, at B and between C and D.For Mmax at B:Examine a section cut just to the left or right of theconcentrated load. M = 110 k2110¿2 = 100 k-ft. 1 - 2 momentFFor Mmax between C and D:Examine the equation of section cut c-c. [ΣFy = 0]4 - 10 k + 20 k - 2 k/t.(x - 20’) - V = 0 -10 k + 20 k - 2x + 40 - V = 0 ‹ V = 50 - 2xBut, Mmax occurs at V = 0. ‹ 0 = 50 - 2x, x = 25’ [ΣMc = 0] 4 at x = 25’ [ΣMc = 0] 4 at x = 25’ +10 k(25’)- 20 k(15’) + 2 k/ft.(5’)(2.5’) + M = 0Mmax = 25 k-ft. (+) momentNote: Beams with one overhang end develop two possiblevalues. ‹ Mcritical = 100 k-ft. at B ( - ) moment

Construct the resulting shear and moment diagrams.

To find examine section (a) left and right of concentratedloads and section (b) at the beginning and end of distributed loads.

FBD at section cut a-a. Section a-a, x = 0 to x = 10’ [ΣFy = 0]V = 10 k ( - ) shear Just right of A, V = 10 k At x = 10’, V = 10 k

FBD at section cut b-b. Section b-b, x = 10’ to x = 20’ Just right of B, ΣFy = 04V = 10 k (+) shear (constant) Just left of C, at x = 20’, ΣFy = 04 and V = 10 k

FBD at section cut c-c. Section c-c, x = 20‘ to x = 30’ Just right of C, ΣFy = 04 V = 10 k ( + ) shear Just left of D, at x = 30’, ΣFy = 04 V = 20 k - 10 k - 2 k/ft.(x - 20)’ V = 50 k - 2x V = 50 k - 2x

Draw shear and moment diagrams for an overhang beam loaded as shown. Determine the critical and Locations and magnitudes. Draw an FBD. Solve for external reactions. Based on intuition, sketch the deflected shape of the beam to assist in determining the signs for moment.

Cut sections a, b, and c between loads and reactions.

Load diagram (FBD).

Loaded beam.

APPLICATION OF STRENGTH IN MATERIALS IN ARCHITECTURE

A shear diagram is a graph in which the abscissa(horizontalreference axis) represents distances along the beam lengthand the ordinates(vertical measure-ments from the ab-scissa) represent the transverse shear at the correspondingbeam sections.

A moment diagram is a graph in which theabscissa represents distances along the beam and the ordi-na-tes represent the bending moment at the correspond-ingsections.Shear and

For the beam to remain in equilibrium, an internal forcesystem must exist within the beam to resist the appliedforces and moments. Stresses and deflections in beams arefunctions of the internal reactions, forces, and moments.

MAPUA INSTITUTE OF TECHNOLOGY

SOLUTION:INTRODUCTION:

MEC32/C2KASAI LIM LIPANAC6 C7 C8 C9 C10LOZANO MORENO

PROBLEM: