×
Log in to StudySoup
Get Full Access to Cal State Fullerton - EGME 421 - Class Notes - Week 2
Join StudySoup
Get Full Access to Cal State Fullerton - EGME 421 - Class Notes - Week 2

Already have an account? Login here
×
Reset your password

CAL STATE FULLERTON / OTHER / EGME 421 / How does axial stress vary from neutral axis?

How does axial stress vary from neutral axis?

How does axial stress vary from neutral axis?

Description

School: California State University - Fullerton
Department: OTHER
Course: Mechanical Design
Professor: Dr. joseph piacenza
Term: Spring 2017
Tags: fundamentals of mechanical design and stress
Cost: 25
Name: EGME 421 week 2
Description: these notes went over stress distribution including bending, torsion, and shear stresses
Uploaded: 02/07/2017
9 Pages 100 Views 1 Unlocks
Reviews


continue stress distribution (week 2)


How does axial stress vary from neutral axis?



Tuesday, January 31, 2017

4:14 PM

normal stresses for beams in bending

. bending stress varies linearly with distance from neutral axis, y

ox=- My

Tis the second-area moment about the z axis

I=12dA


What is the second moment of area about any axis?



transverse shear stress

unglued slabs unloaded

unglued slabs loaded

Sad Sour

*

*

MANNA

***

glued slabs unloaded

glued slabs loaded

SER

T = X SATA If you want to learn more check out What is the meaning of ionic in intermolecular interactions?

Q

first moment

of area

circh

how

Than = av


What is longitudinal shear stress?



Tmax at the neutral axis Y-zero at the outer

Surface A often very small compared to

Torsion and Scending

bending moment diagram

- a simply supported rectangular bar in bending

Pf Don't forget about the age old question of What does tr stand for in economics?

T Pa

O -V=0

AM

- Mmax

Combined loading - round bar in bending and torsion example: Porsche 911 rear torsion bar Don't forget about the age old question of What is the focus of comparative criminology?

solid bar

w

Study SO

top surface A max stress due to bending

Omax = MC

point B max stress due to bending

max shear stress due to torsion

Tmax = 7 -

Omax=0

transverse shear

max shear stress due to torsion

Trending 0

transverse shear

I senting

Thursday, February 2, 2017

example (3.8): figure shows a crank loaded by a force F=300 lbf that causes twisting and bending of a 3/4 in diameter shaft fixed to a support at the origin of the reference system. In actuality, the support may be an inertia that we wish to rotate, but for the purposes of a stress analysis we can consider this a statics problem We also discuss several other topics like Who were the first hominids?

a) draw separate free body diagrams of the shaft AB and the arm BC, and compute the value of all

forces, moments, and torques that act. Label the directions of the coordinate axes on these diagrams compute the maxima of the torsional stress and the bending stress in the arm BC and indicate If you want to learn more check out What is the meaning of computer forensics?
Don't forget about the age old question of What is the meaning of public issues in sociology?

where these act. c) locate a stress element on the top surface of the shaft at A, and calculate all the stress

components that act upon this element d) determine the maximum normal and shear stresses at A

(@StudySoup

» Study Soup

Stud Sour

Fig. 3-22

Studysoup

StudySoup

StudySoup

FBD

shaft AB

-

SENSO

FBD

arm BC

shaft BC Point Bi

M = 300 lbf · 4 in = 1200 lbf -in

Ti=300 lbf - 1.5in = 450 lbf in T = -300 lbf •1.5=-450 lbf - in

Point C:

| Shaft AB

Point B: Ta=-300 lff • 4in=- 1200 lbf "in

- Ma =-30016f. 1.5in = -450 lbf in Point A: Ma = 300 lf (1.5+5) in= 1950 lbf in

To =300 lbf Hin = 1200 lbf oin 6 compute stress on arm Bc book at Bending:

Tunda Me - o

coto

n

CM tha

(1200) (25)(r.as)?

8.4 ksi

Is bha

hos oss) 1.s kial U

19.4 ksi

© locate max. stress at point A

Granding Me 3 24

= 32(1950)

--- 147.1 ksi

T(175) max=-Tr=-16T

(1900), -14.5 ki

Page Expired
5off
It looks like your free minutes have expired! Lucky for you we have all the content you need, just sign up here