chapter 9 combined stresses
Download
Skip this Video
Download Presentation
Chapter 9 Combined Stresses

Loading in 2 Seconds...

play fullscreen
1 / 103

Chapter 9 Combined Stresses - PowerPoint PPT Presentation


  • 181 Views
  • Uploaded on

Chapter 9 Combined Stresses. 9-1 Introduction. Basic types of loading: axial, torsional and flexural Stress formulas: Axial loading - Torsional loading - Flexural loading -. 9-2 Combined Axial & Flexural Loads.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Chapter 9 Combined Stresses' - amity-ramos


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
9 1 introduction
9-1 Introduction
  • Basic types of loading: axial, torsional and flexural
  • Stress formulas:

Axial loading -

Torsional loading -

Flexural loading -

slide11

For stiff members the formula is appropriate

For long slender members or columns, the effect of P-d is significant

slide14

Hw10

sallow

B

D2

D1

Fig. P-908

ค่า z1-z6 ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้

46z1z2z3z4z5z6

D1=(1+z1) in. D2 = D1(1+z2) in.

I1-1=1000(1+z3) in4 Area=10(1+z4) in2

B =10(1+z5) in. sallow=10(1+z6) ksi.

หมายเหตุD2= D1(1+z2) in.

เพื่อให้หน้าตัดมีประสิทธิภาพดีในการรับหน่วยแรง

slide15

Hw11

L4

L2

L3

b

h

L1

ค่า z1-z6 ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้

46z1z2z3z4z5z6

L1= (1+z1) in. L2 = (1+z2) in.

L3= (1+z3) in. L4 = (1+z4) in.

b = 0.2(1+z5) in. h = b(1+z6) in.

P =(1+z5) kips. F =(1+z6) kips.

หมายเหตุh = b(1+z6) in.

เพื่อให้คานมีความลึกไม่น้อยกว่าความกว้างเสมอ

slide17

The maximum eccentricity to avoid tension

The general case:

The position of neutral axis (line of zero stress)

That is in designing of masonry or other structures weak in tension, the resultant load should fall in the middle third of the section.

slide19

918

A compressive load P= 12 kips is applied, as in Fig. 9-8a, at a point 1 in. to the right and 2 in. above the centroid of a rectangular section for which h=10 in. and b=6 in. Compute the stress at each corner and the location of the neutral axis. Illustrate the answers with a sketch similar to Fig. 9-8b.

slide21

921

Calcualte and sketch the kern of a W360 X 122 section.

9 5 stress at a point
9-5 Stress at A Point

Stress at a point really defines the uniform stress distributed over a differential area.

slide25

The most general state of stress at a point may be represented by 6 components,

symmetry

state of stress เมื่อแสดงด้วยระบบโคออร์ดิเนต (xyz)

symmetry

state of stressเมื่อแสดงด้วยระบบโคออร์ดิเนต (xyz)

slide26

Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by

  • State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate.
  • State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface not subjected to an external force.
slide27

Plane Stress

Two methods to compute the maximum stresses i.e.,

  • Analytical approach
  • Using of Mohr’s circle
slide32

Find maximum or minimum s differentiating Eq.(9-5) w.r.t. q and setting the derivative equal to zero

Find maximum or minimum t differentiating Eq.(9-6) w.r.t. q and setting the derivative equal to zero

Eq.(9-5)

Eq.(9-6)

slide33

At zero shearing stress t = 0

Eq.(9-5)

Eq.(9-6)

ซึ่งเป็นมุมเดียวกับสมการ Eq.(9-7)ดังนั้น ค่า maximum or minimumsจะเกิดขึ้นเมื่อ t = 0

slide34

มุมqและ qs ต่างกัน 45O

Maximum or minimum t

Maximum or minimum s (Principal stresses)

9 7 variation of stress at a point mohr s circle
9-7 Variation of Stress at A Point: Mohr’s Circle

Otto Mohr (1882)

Eq.(9-5)

Eq.(9-6)

Eq.(a)2 + Eq.(b)2

slide40

x-axis

y-axis

Rule for Applying Mohr Circle to Combined Stresses

slide41

x-axis

C

y-axis

slide42

x-axis

n-axis

R

2q

q

C

y-axis

slide43

x-axis

n-axis

R

2q

q

C

y-axis

slide44

R

2q1

x-axis

C

y-axis

2q2

slide46

y-axis

R

C

2q1

x-axis

slide47

y-axis

R

60o

C

x-axis

45o

9 8 absolute maximum shearing stress

s2

s1

s2

s1

s1

s2

9-8 Absolute Maximum Shearing Stress

Mohr’s circle: Rotation around z-axis

slide51

s2

s1

Mohr’s circle: Rotation around x-axis

Mohr’s circle: Rotation around y-axis

slide52

s2

s1

s2

s1

slide53

s2

s1

Mohr’s circles for plane stress

Absolute maximum shearing stress for plane stress is equal to the largest of the following three values
slide54

s2

s1

s3

z

Mohr’s circles for general state of stress

Absolute maximum shearing stress for general state of stress is equal to the largest of the following three values
slide55

20

50

Maximum in-plane shearing stress =

Absolute maximum shearing stress is the largest of

slide56

20

50

Ex.

Maximum in-plane shearing stress =

Absolute maximum shearing stress is the largest of

slide57

the figure

( สำหรับข้อนี้ให้คำนวณ ค่าabsolute maximum shearing stress ด้วยโดยกำหนดให้ sz=0 )

Hw17

ค่าz1-z3ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้46xxxz1z2z3

9 9 application of mohr s circle to combined loadings

Combined stresses

Mohr’s Circle

Design Criteria,

y-axis

Principal stresses and, Maximum shearing stress

x-axis

9-9 Application of Mohr’s Circle to Combined Loadings

Combined Loadings (axial, torsional, flexural)

torsional failure modes

Ductile materials generally fail in shear.Brittle materials are weaker in tension than shear.

  • A ductile specimen breaks along a plane of maximum shear
  • A brittle specimen breaks along planes perpendicular to s1
Torsional Failure Modes

45o

slide61

Stress Trajectories for Torsion

Stress Trajectories: lines of principal stress direction but of variable stress intensity

slide62

y-axis

x-axis

Mohr’s Circle

Stress Trajectories for Beam

slide70

BMzD

TMD

BMyD

slide71

E

D

C

B

BMyD

A

BMzD

TMD

|M|

D

A

B

E

C

Cross section of solid shaft

and the resultant moment

slide72

At section C

BMyD

BMzD

From Prob. 951 and this problem.

Mohr’s Circle

y-axis

TMD

At section D

|M|

D

A

B

E

C

x-axis

slide79

Hw18

ค่า z1-z5 ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้46xz1z2z3z4z5

L1= 4(1+z1) in. L2 = 4(1+z2) in.

L3= 4(1+z3) in. L4 = 4(1+z4) in.

D= 4(1+z5) in.

slide80

Hw19

Also find the maximum shearing stress at point A. Show your results on a complete sketch of a differential element.

ค่า z1-z4 ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้46xxz1z2z3z4

L= 0.4(1+z1)m. P= 4(1+z2)kN

H= 40(1+z3)mm. W= 40(1+z4)mm

slide89

Eq.(9-5)

Eq.(9-6)

slide99

จงพิสูจน์ สมการ (9-19) (9-20) ด้วยภาษาของตัวเอง

Hw20a

Hw20b

Hw21

ค่า z1-z3 ได้จากเลขประจำตัวนิสิต ดังต่อไปนี้46xxxz1z2z3

ea= 100(1+z1)eb= -100(1+z2)

ec= 100(1+z3)

slide100

ปริมาณทางPhysics สามารถแทนด้วยTensor

Order 0 = zero order Tensor (Scalar) – Magnitude (มวล, ความหนาแน่น)

Order 1 = first order Tensor (Vector) – Magnitude, Direction (ความเร็ว, แรง)

Order 2 = second order Tensor – Magnitudes, Directions(stress, strain)

… Higher order ….

ปริมาณทางPhysics ไม่เปลี่ยนแปลงไปตามระบบโคออร์ดิเนตที่ใช้ในการวัด

slide101

ปริมาณทางPhysics ไม่เปลี่ยนแปลงไปตามระบบโคออร์ดิเนตที่ใช้ในการวัด

แรงยังคงมีขนาดและทิศทางเท่าเดิม ไม่ว่าจะแสดง componentของเวคเตอร์ด้วยระบบโคออร์ดิเนตอื่น

สถานะของหน่วยแรง (state of stress)ยังคงมีคุณสมบัติเหมือนเดิม ไม่ว่าจะแสดงด้วยระบบโคออร์ดิเนตอื่น

ad