pdf lec8 transverse shear
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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