Download The flexure formula is valid oly when bending is about a ... and more Exercises Mechanics in PDF only on Docsity!
CE560 Advanced Mechanics of Solids
Non-symmetric Pure Bending
The exure formula is valid oly when b ending is ab out a principal axis. When this is not the case, then the moment can b e divided into comp onents in the direction of principal axes, and their results summed using the principle of sup erp osition.
EXAMPLE:
Given a b eam with the cross-section shown and sub jected to a pure b ending moment ab out a vertical axis. Determine the maximum and minimum axial stresses. (Note: units of length and force can b e in any consitant set of units).
Step 1: Determine the centroid.
Section Area X Y XArea YArea 1 16 1 4 16 64 2 24 8 1 192 24 SUM 40 208 88
y
x
Xc = 20840 = 5 : 2
Yc = 8840 = 2 : 2
Step 2: Determine Moments of Inertia.
x
y
-4.
Section Area Ixx Iy y Ixy X Y Ixx Iy y Ixy (A) +AY 2 AX 2 +AX Y 1 16 85.33 5.33 0.0 -4.2 1.8 287.57 137.17 -121. 2 24 8.00 288. 0.0 2.8 -1.2 476.16 42.567 -80. SUM 763.8 179.7 -201.
Step 3: Determine Principal Moments of Inertia by Mohr's Circle
y(763.7. 201.6)
x(179.6, -201.6)
C
Radius = 354.
C = 471.
I = 826.
I = 116.
I
I , I
xy
xx yy
2θ = 34.62o
y’
x’
Step 6: Determine the Orientation of the Neutral Axis.
The neutral axis is the lo cus of p oints for which the axial stress is zero. It is the axis ab out which the cross-section rotates. We nd its equation by setting the axial stress equal to zero. Hence:
2 :541(10 ^3 )y 0 1 :155(10 ^3 )x^0 = 0
or simply, y 0 = : 455 x^0
The angle which this line makes with the x^0 axis is:
= atan( :455) = 24 : 5 deg
x
y
y’
x’
o
M
Neutral Axis
o
Step 7: Determine Maximum Tensile and Compressive Stress
The maximum values of tensile and compress stress o ccur at the farthest p oints each side of neutral axis. These p oints are A and B as shown.
x
y
y’
B x’
A
N.A.
o
The x and y co ordinates of these p oints are:
Point x y A 8 : 8 0 : 2 B 5 : 2 2 : 2
and by using the matrix for co ordinate rotations, the x^0 and y 0 co ordinates are:
x^0 A y (^0) A
cos(17:3) sin(17:3) sin(17:3) cos(17:3)
and (
x^0 B y (^0) B
cos (17:3) sin(17:3) sin(17:3) cos (17:3)
Thus, the axial stresses at these p oints are:
A = 2 :541(10 ^3 )(2:43) 1 :155(10 ^3 )(8:46)
= 15 : 9 units of stress in Compression
and B = 2 :541(10 ^3 )( 3 :65) 1 :155(10 ^3 )( 4 :31)
= +14: 3 units of stress in Tensions
x
y
B
A
y’
x’
14.3 (T)
15.9 (C)
flexural stress
N.A.
