Combined Stresses –Combined Normal Stresses Axial(P)/Bending (M) Page A 1Page 531 - 532

Combined Stresses –Combined Normal Stresses Axial(P)/Bending (M) Page A 2

Calculate tensile stress

Calculate bending stress

Find total

708

Combined Stresses –Combined Normal Stresses Axial(P)/Bending (M) Page A 3

10.6 page 588 Tension in blade is 125 N Draw free-body of top beam

Find compressive stress

Find bending stress

Find combined worst case

Combined Stresses –Combined Normal Stresses Axial(P)/Bending (M) Page A 4

10.16 page 590 EF Tensile 54 kBack to back steel angles, ASTM A360.6 yield strength Pg 696 angles

Try 4 x 3 x 1/4

Free-body of EF

Shear Diagram

Moment Diagram

Populate

σ= P2 A

+ M2S

Estimate need A for axial only

Estimate needed S for bending only

Combined Stresses – Combined Normal and Torsional Shear Stress Page B 1

10-28 page 592-3Choose standard pipeHold Maximum Shear stress to 8000psi

**Maximum Shear Stress Theory of Failure page 546 (MAJOR CRITERIA)

τ max=√( σ2 )2

+τ2<S ys

Member fails when the maximum shear stress exceeds the yield strength of the material in shear.

Identify worst case element

Find M on element

Find T on element

Select pipe page 710 Find ZP needed

Simplified for Torque and Bending

σ=MSS=π D3

32

τ= TZP

Z P=π D3

16

ZP = 2S

τ max=√(σ2 )2

+τ2<S ys

τ max=√( M2 S )2

+( TZP )2

= 1ZP

√M 2+T 2<S ys

Equivalent Torque=T e=√M 2+T 2

with stress concentration=T e=√ (K tBM )2+(K tTT )2

τ max=T e

ZP

Combined Stresses – Rotating Shafts, Page C 1

Page 546 - 551

Combined Stresses – Rotating Shafts, Page C 2

10.32 page 594, shaft diameter 1.75 inFind maximum shear stress

Sketch torques at each pulley

Find y force at bearings

Sketch shear and moment diagrams

Find τmax at C

τ max=√(σ2 )2

+τ2

Combined Stresses – Axial/Direct Shear Page D 1Bolt in tension σ and connection in pure shear τBoth stresses evenly distributed over cross section

xy

τ max=√( σ2 )2

+τ2

10-40 page 595Machine screw diameter 48 mm, pitch 5.0mm page 694Tensile stress 120 MPa on thread tensile areaNon treaded area pure shear force of 80kN.Find the maximum shear stress

Find force Find τ

Find combined

Find σ at element

Combined Stresses – Bending/Vertical Shear Page E 1

Select worst case combined V and M

Find σ at y

Find τ at y

Combine to get τmax

Shaft D = 50mm

Noncircular Shapes Page F 1

Page 227 use Roark for other shapes

10-46 page 595 Triangle 50mm sidesAxial force 115 kNTorque 775 NmFind maximum shear stress.

Find σ

Find τ

Find τmax

General Combined Stresses Page G 1

Sign convention and naming τxy on x face in y direction

∑ Fu=0 collect terms page 553Normal stress in u direction

σ u=12 (σ x+σ y )+ 1

2 (σx−σ y ) cos2φ−τ xy sin2φ

Shear stress, uv

τuv=−12 (σ x−σ y ) sin 2φ−τ xycos2φ

Differentiate to find angle φ that σ maximizes (page 555)σ1 and σ2 principal stresses

φ=12

tan−1[ −τ xy12 (σ x−σ y ) ]→ tan 2φ=

−τ xy12 (σx−σ y )

σ max=σ1=12 (σ x+σ y )+√[1

2 (σ x−σ y) ]2

+ τ xy2

σ min=σ2=12 (σx+σ y )−√ [1

2 (σ x−σ y )]2

+τ xy2

Differentiate to find angle φ that τ maximizes (page 556)

φ=12

tan−1[ 12 (σ x−σ y )

−τ xy ]→ tan 2φ=

12 (σx−σ y )

−τxy

τ max=∓ √[ 12 (σx−σ y )]

2

+τ xy2

Force = stress x areaOriginal area hxh

Substitute into σu and τuv

Shear

General Combined Stresses Page G 2

WHAT THE ANGLE MEANS

10.60 page 597σx = 20 ksi σy -5 ksi τxy 10 ksi CCWcalculate principal stressesmaximum shear stressaverage normal stressangle φ and φ’

Angle from x axis to principal stress

tan2φ=−τ xy

12 (σ x−σ y )

get quadrant ¿ signs¿

σ 1=12 (σ x+σ y )+√ [1

2 (σx−σ y )]2

+ τxy2

σ 1=12

(20ksi−5ksi )+√[ 12

(20ksi+5 ksi )]2

+(−10ksi)2

σ 1=7.5ksi+16.0 ksi=23.5kis

σ 2=7.5ksi−16.0 ksi=−8.5ksi

τ max=∓16.0 ksi

σ avg=σ1+σ 2

2=23.5ksi−8.5ksi

2=7.5ksi

φ=12

tan−1[ −τ xy12 (σ x−σ y ) ]=1

2tan−1¿¿

φ=19.3deg ccw from x

φ '=900−2φ2

=900−2(−19.30)

2=64.3deg

General Combined Stresses Page G 3

Normal stress in u direction

σ u=12 (σ x+σ y )+ 1

2 (σx−σ y ) cos2φ−τ xy sin2φ

σ u=12 (300+(−100))+ 1

2¿

σ u=130.7MPa

Shear stress, uv

τuv=−12 (σ x−σ y ) sin 2φ−τ xycos2φ

τuv=−12 (300−(−100)) sin 2(30)−80 cos 2(30)

τuv=213.2MPa

10-96 page 598σx = 300 MPa, σy = -100 MPa, τxy = 80 MPa cw30 deg ccw

General Combined Stresses – Mohr’s Circle Page H 1

10.60 page 597σx = 20 ksi σy -5 ksi τxy 10 ksi CCWUse Mohr’s circle to find principal stressesmaximum shear stressaverage normal stressangle φ and φ’

Plot σx and τxy, σy and τyx

Use circle geometry instead of equations

CW

20,10ccw

-5,10cw

2φ

General Combined Stresses – Mohr’s Circle Page H 2

Von Mises criteria used for ductile materials – based on principal stresses. http://homepages.uc.edu/~caldwelm/Courses/strength_theories.pdf

10-92 page 598σx = -840 kPa σy -335 kPa τxy 120 kPa CCWUse Mohr’s circle to find principal stressesmaximum shear stress (enhanced)average normal stressangle φ and φ’

σ e=√22 √(σ1−σ2)

2+(σ2−σ3)2+(σ3−σ1)

2=S y

N

General Combined Stresses – Mohr’s Circle Page H 3

General Combined Stresses – Mohr’s Circle Page H 4

General Combined Stresses – Mohr’s Circle Page H 5