Lecture 18 - Mechanical and Aerospace Engineering

Transcript Lecture 18 - Mechanical and Aerospace Engineering

Mechanics of Materials – MAE 243 (Section 002) Spring 2008

Dr. Konstantinos A. Sierros

Problem 4.3-14

The simply-supported beam

ABCD

is loaded by a weight W = 27 kN through the arrangement shown in the figure. The cable passes over a small frictionless pulley at

and is attached at

to the end of the vertical arm. Calculate the axial force

, shear force

, and bending moment M at section C, which is just to the left of the vertical arm.

(

Note:

Disregard the widths of the beam and vertical arm and use Center line dimensions when making calculations.)

Problem 4.5-14

The cantilever beam

shown in the figure is subjected to a uniform load acting throughout one-half of its length and a concentrated load acting at the free end.

Draw the shear-force and bending-moment diagrams for this beam.

5.5: Normal stresses in beams (Linearly elastic materials)

• Since longitudinal elements of a beam are subjected only to tension/compression, we can use the

stress-strain curve

of the material to determine the stresses from the strains • The most common stress-strain relationship encountered in engineering is the equation for a

linearly elastic material

Resultant of the normal stresses 1) A force acting in the x-direction 2) A bending moment acting about the z-axis

FIG. 5-9

Normal stresses in a beam of linearly elastic material: (a) side view of beam showing distribution of normal stresses, and (b) cross section of beam showing the z axis as the neutral axis of the cross section

5.5: Location of neutral axis

• Consider an element of area dA in the cross-section. The element is located at distance y from the neutral axis • This equation states that the z-axis must pass through the centroid of the cross-section The neutral axis (z-axis) passes through the centroid of the cross-sectional area when the material follows Hooke’s law and there is no axial force acting on the cross-section

FIG. 5-9

5.5: Moment curvature relationship

•

Moment of inertia

of the cross-sectional area with respect to the z-axis •The

moment-curvature equation

shows that the curvature is directly proportional to the bending moment M and inversely proportional to the quantity EI, which is called the

flexural rigidity

Positive bending moment produces positive curvature and a negative bending moment produces negative curvature

FIG. 5-10

Relationships between signs of bending moments and signs of curvatures

5.5: Flexure formula

• We can determine the stresses in terms of the bending moment • This equation is called the

flexure formula

and shows that the stresses are directly proportional to the bending moment M and inversely proportional to the moment of inertia I of the cross-section • Stresses that are calculated from the

flexure formula

are called

bending stresses

flexural stresses

5.5: Maximum stresses at a cross section

• The maximum tensile and compressive bending stresses acting at any given cross-section occur at points located furthest for the neutral axis • The corresponding maximum normal stresses σ 1 and σ 2 (

from the flexure formula

) section moduli

FIG. 5-11

Relationships between signs of bending moments and directions of normal stresses: (a) positive bending moment, and (b) negative bending moment

5.5:Doubly symmetric shapes

• If the cross-section of a beam is symmetric with respect to the z-axis as well as the y-axis then we have or For a beam of rectangular cross-section with width b and height h

FIG. 5-12

Doubly symmetric cross-sectional shapes