PDF Lec8 Transverse Shear

Lecture 1 1

Chap 7 Transverse Shear

Objective

Transverse Shear Formula

Transverse Shear

Lecture 1 2

I

My

Transverse Shear

Shear Formula

Lecture 1 3

It

VQ

: the shear stress in the member at the point located a

distance y’ from the neutral axis. This stress is assumed to be constant and therefore averaged across the width t of the member

V: the internal resultant shear force.

I: the total moment of inertia the entire cross sectional

area calculated about the neutral axis

t: the width of the member’s cross sectional area,

measured at the point is to be determined

Q: where A’ is the area of the top (or bottom) portion of

the member cross-sectional area, above (or below) the section plane where t is measured, and is the distance from the neutral axis to the centroid of A’

'Ay

y

Shear Formula

Lecture 1 4

It

VQ

Neutral Axis

I of the total cross sectional area

V = 1kN

Shear Formula

Lecture 1 5

It

VQ

mm

AA

AyAyy

43.41

)50(20)40(10

)50)(20(50)40)(10(20

21

2211

Neutral Axis

45

23

23

222_111_

)10(438.3

])43.4150)(50(2012

)20(50[])2043.41)(40(10

12

)40(10[

][][

mm

I

dAIdAII prfprf

I of the total cross sectional area

V = 1kN

Shear Formula

Lecture 1 6

Shear at yellow line

V= 1000 N Q=yA=(9.285)(18.57)(50) I = I total t = 50 mm

MPa

It

VQ

502.0

50)10(438.3

)50)(57.18)(285.9(10005

Shear stress

Shear Formula

Lecture 1 7

Shear at yellow line

V= 1000 N Q=yA=13.57(10)(50) I = I total t = 50 mm

MPa

It

VQ

395.0

50)10(438.3

)50)(10)(57.13(10005

Shear stress

Lecture 1 8

Shear at yellow line

V= 1000 N Q=yA=21.43(10)(40) I = I total

MPa

It

VQ

623.0

50)10(438.3

)50)(10.)(43.21(10005

Shear stress above yellow (t = 50 mm

MPa

It

VQ

117.3

10)10(438.3

)50)(10.)(43.21(10005

Shear stress bellow yellow (t = 10mm)

Lecture 1 9

Determine the shear stress at point A and B

Lecture 1 10

What do expect the shear stress at the top and bottom line?

For the cross sectional areas that have the symmetrical shape below and top of the neutral axis. Where is the maximum shear stress?

Solve it F7-3 pp 373

Lecture 1 11

Determine the absolute maximum shear stress developed in the beam

Lecture 1 12

Vmax at the neutral axis V= 22.5 kN Q=yA=37.5(75)(75)

46

3

)10(09.21

12

)150(75

mm

I

MPa

It

VQ

0.3

)75()10(09.21

)75)(75)(5.37)(22500(6max

Solve it 7-9/10 pp 375

Lecture 1 13

7-9) Determine the largest shear force V that the member can sustain if the allowable shear stress is allow = 56 MPa 7-10) If the applied shear force V=90 kN, determine the maximum shear stress in the member

Lecture 1 14

It

VQ

mm

AA

AyAyy

833.45

)50(75)75(125

)50)(75(25)75)(125(5.37

21

2211

Neutral Axis

Moment of Area

46

23

23

222_121_

)10(6367.2

])25833.45)(50(7512

)50(75[])5.37833.45)(125(75

12

)75(125[

][][

mm

I

dAIdAII prfprf

Lecture 1 15

Vmax at the neutral axis V= V N Q=yA=22.9165(2)(25)(45.833) I = 2.636x106 mm4

MPaV

V

It

VQ

4

6max

)10(985.3

50)10(636.2

)833.45)(50)(9165.22(

kNV

V

all

53.140

56)10(985.3 4

max

Since max = all

Lecture 1 16

Vmax at the neutral axis V= 90,000N Q=yA=22.9165(2)(25)(45.833) I = 2.636x106 mm4

MPa

It

VQ

87.35

50)10(636.2

)833.45)(50)(9165.22(900006max

HW

7-6, pp 374,

7-12 pp 375,

7-21, 7-23 pp 376

Lecture 1 17