Transcript Document

Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.04
SOLUTION:
• Determine the shear force per unit
length along each edge of the upper
plank.


VQ 600 lb  4.22 in 3
lb
q


92
.
3
I
in
27.42 in 4
q
lb
 46.15
2
in
 edge force per unit length
f 
For the upper plank,
Q  Ay  0.75in. 3 in .1.875 in .
 4.22 in 3
For the overall beam cross-section,
1 4.5 in   1 3 in 
I  12
12
4
4
 27.42 in 4
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
• Based on the spacing between nails,
determine the shear force in each
nail.
lb 

F  f    46.15 1.75 in 
in 

F  80.8 lb
6-1
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• Consider a segment of a wide-flange
beam subjected to the vertical shear V.
• The longitudinal shear force on the
element is
H 
VQ
x
I
• The corresponding shear stress is
 zx   xz 
H VQ

t x It
• Previously found a similar expression
for the shearing stress in the web
 xy 
VQ
It
• NOTE:  xy  0
 xz  0
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
in the flanges
in the web
6-2
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• The variation of shear flow across the
section depends only on the variation of
the first moment.
q t 
VQ
I
• For a box beam, q grows smoothly from
zero at A to a maximum at C and C’ and
then decreases back to zero at E.
• The sense of q in the horizontal portions
of the section may be deduced from the
sense in the vertical portions or the
sense of the shear V.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-3
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Shearing Stresses in Thin-Walled Members
• For a wide-flange beam, the shear flow
increases symmetrically from zero at A
and A’, reaches a maximum at C and
then decreases to zero at E and E’.
• The continuity of the variation in q and
the merging of q from section branches
suggests an analogy to fluid flow.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-4
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
I
M

 Y  maximum elastic moment
• Recall: Y
c
• For M = PL < MY , the normal stress does
not exceed the yield stress anywhere along
the beam.
• For PL > MY , yield is initiated at B and B’.
For an elastoplastic material, the half-thickness
of the elastic core is found from
 1 yY2 
3
Px  M Y 1  2 
 3c 
2


• The section becomes fully plastic (yY = 0) at
the wall when
3
PL  M Y  M p
2
• Maximum load which the beam can support is
Pmax 
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Mp
L
6-5
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• Preceding discussion was based on
normal stresses only
• Consider horizontal shear force on an
element within the plastic zone,
H   C   D dA   Y   Y dA  0
Therefore, the shear stress is zero in the
plastic zone.
• Shear load is carried by the elastic core,
3 P 
 xy 
1
2 A 
 max 
y 2 
where A  2byY
2
yY 
3P
2 A
• As A’ decreases, max increases and
may exceed Y
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-6
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 6.3
SOLUTION:
• For the shaded area,
Q  4.31in 0.770 in 4.815 in 
 15.98 in 3
• The shear stress at a,
Knowing that the vertical shear is 50
kips in a W10x68 rolled-steel beam,
determine the horizontal shearing
stress in the top flange at the point a.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.



VQ 50 kips  15.98 in 3


It
394 in 4 0.770 in 

  2.63 ksi
6-7
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Unsymmetric Loading of Thin-Walled Members
• Beam loaded in a vertical plane
of symmetry deforms in the
symmetry plane without
twisting.
x  
My
I
 ave 
VQ
It
• Beam without a vertical plane
of symmetry bends and twists
under loading.
x  
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
My
I
 ave 
VQ
It
6-8
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Unsymmetric Loading of Thin-Walled Members
• If the shear load is applied such that the beam
does not twist, then the shear stress distribution
satisfies
D
B
E
VQ
 ave 
V   q ds F   q ds    q ds   F 
It
B
A
D
• F and F’ indicate a couple Fh and the need for
the application of a torque as well as the shear
load.
F h  Ve
• When the force P is applied at a distance e to the
left of the web centerline, the member bends in a
vertical plane without twisting.
• The point O is referred to as the shear center of
the beam section.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
6-9
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.05
• Determine the location for the shear center of the
channel section with b = 4 in., h = 6 in., and t = 0.15 in.
e
Fh
I
• where
b
b VQ
Vb h
F   q ds  
ds   st ds
I0 2
0
0 I
Vthb2

4I
2
1 3
1 3
h 
I  I web  2 I flange  th  2  bt  bt   
12
 2  
12
1 th2 6b  h 
 12
• Combining,
e
b
h
2
3b

4 in.
6 in .
2
34 in .
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
e  1.6 in .
6 - 10
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 6.06
• Determine the shear stress distribution for
V = 2.5 kips.

q VQ

t
It
• Shearing stresses in the flanges,
VQ V
h Vh
 st   s
It
It
2 2I
Vhb
6Vb
B 

2 1 th2 6b  h  th6b  h 

12 

62.5 kips 4 in 
 2.22 ksi
0.15 in 6 in 6  4 in  6 in 
• Shearing stress in the web,
 max
 
1
VQ V 8 ht 4b  h  3V 4b  h 



1 th2 6b  h t
It
2th6b  h 
12

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
32.5 kips 4  4 in  6 in 
 3.06 ksi
20.15 in 6 in 6  4 in  6 in 
6 - 11