lecture slides - week 8-2
TRANSCRIPT
-
8/12/2019 Lecture Slides - Week 8-2
1/9
M(x)+dM(x)
top=0
The Shear Formula
t(y)
dA=t(y)dy
A
yx (y)
xy(x)
Force Equilibrium =0+
-
8/12/2019 Lecture Slides - Week 8-2
2/9
The Shear Formula
t(y)
M(x)+dM(x)
top=0
dx
y
z
x
V
V: the internal resultant shear force at the point
I: the moment of inertia of entirecross-sectional area
t: the width of the members cross-sectional area at the point
Q: , where A is the top portion of the members cross sectional area, and y
is the distance to centroid of A measured from neutral axis.
A
-
8/12/2019 Lecture Slides - Week 8-2
3/9
Shear Stresses Distribution in Beams
The Shear Formula
(y=0)
-
8/12/2019 Lecture Slides - Week 8-2
4/9
Analysing Bending Test
N.A.
Shear Stress
M
LP/4
V
P/2
-P/2
PL x t x ha
a
Step 1. Shear and moment diagrams
a. Equilibrium condition (support reactions)
b. Draw shear and moment diagram
;
c. Maximum shear force and bending moment
Tensile Stress
-
8/12/2019 Lecture Slides - Week 8-2
5/9
Step 2. Maximum normal and shear stresses in the beam
Analysing Bending Test
a
a
PL x t x h
a. Neutral axis
b. Inertial moment
c. Stress distribution:
d. Maximum normal stress
e. Maximum shear stress
t
h
N. A.
-
8/12/2019 Lecture Slides - Week 8-2
6/9
-
8/12/2019 Lecture Slides - Week 8-2
7/9
300 mm
20 mm
20 mm
20 mm
200 mm
1. Neutral axis
2. Moment of inertia: I= 2.5 x 10-4m4
N.A.
A
B
C
Shear Stresses Distribution in Beams
3. Shear stress distribution
B
Point A: A= 0
Point B:
Point B:
QB= QB
= 0.16 x (0.02 x 0.2) = 6.4 x 10-4m4
= 2.56 MPa
= 0.256 MPa
7-2 If the wide-flange beam is subjected to a shear of V = 20 kN, (a)plot the shear-stress
distribution acting over the cross-sectional area, and (b) determine the maximum shearstress in the beam. (p. 374)
-
8/12/2019 Lecture Slides - Week 8-2
8/9
1. Neutral axis
2. Moment of inertia: I= 2.5 x 10-4m4
Shear Stresses Distribution in Beams
3. Shear stress distribution 300 mm
20 mm
20 mm
20 mm
200 mm
N.A.
A
B
C
B
Point A: A= 0
Point B:
Point B:
B= 0.256 MPa
Point C:
= [0.16 x (0.02 x 0.2) + 0.075 x (0.02 x 0.15) = 8.65 x 10 -4m4
B= 2.56 MPa
= 3.46 MPa
7-2 If the wide-flange beam is subjected to a shear of V = 20 kN, (a)plot the shear-stress
distribution acting over the cross-sectional area, and (b) determine the maximum shear stressin the beam. (p. 374)
-
8/12/2019 Lecture Slides - Week 8-2
9/9
Shear Stresses Distribution in Beams
Point A: A= 0
Point B:
Point B:
B= 0.256 MPa
Point C:
B= 2.56 MPa
C= 3.46 MPa
a) shear-stress distribution
b) The maximum shear stress
C= 3.46 MPa
A
B B
C
0.256 MPa
2.56 MPa
3.46 MPa
7-2 If the wide-flange beam is subjected to a shear of V = 20 kN, (a)plot the shear-stress
distribution acting over the cross-sectional area, and (b) determine the maximum shear stressin the beam. (p. 374)
300 mm
20 mm
20 mm
20 mm
200 mm
N.A.
A
B
C
B