MEMBER DIFFERENT LOADINGS
(push/pull)
AXIAL
SHEAR
(translates)
TORSIONAL
(twists)
BENDING
(curves)
BENDING
(curves)
BENDING
Consider a beam under a uniform load. By statics, it is clear that RH = 0. By symmetry, R1 = R2 .
+ Fy = 0 = -wL + R1 + R2 R1 = R2 =wL/2
If we cut the beam at "x", there must be an internal shear and moment for equilibrium.
What are the internal shear and the internal moment??
+ Fy = 0 = -wx - V + R1
V = -wx + R1 = w(L/2-x) + MV = 0 = -(wx2)/2 - M + R1x
M = w(x/2)(L x)
SIGN CONVENTION FOR INTERNAL FORCES
M T N
V
M
+
V N
T
BASIC ASSUMPTIONS
OVER A SMALL LENGTH, THE BEAM BENDS INTO A CIRCULAR ARC.
CONSIDER A STRAIGHT BEAM WHICH IS THEN BENT BY A POSITIVE MOMENT AT EACH END
SIDES OF THE ELEMENT REMAIN STRAIGHT PLANE SECTIONS REMAIN PLANE.
d
SIDES REMAIN STRAIGHT. IS LINEAR WITH RESPECT TO y.
e=d/L THEREFORE e IS LINEAR
WITH RESPECT TO y
SIDES OF THE ELEMENT REMAIN STRAIGHT PLANE SECTIONS REMAIN PLANE.
CORNERS REMAIN SQUARE SHEAR DEFORMATIONS ARE SMALL NEGLECT SHEAR DEFORMATIONS
SIDES OF THE ELEMENT REMAIN STRAIGHT PLANE SECTIONS REMAIN PLANE. CORNERS REMAIN SQUARE
1) HORIZONTAL FIBERS CHANGE LENGTH 2) ASSUME STRAIN IN LONGITUDINAL DIRECTION ONLY 3) ONE DIMENSIONAL STRESS/STRAIN FIELD (SLENDER MEMBERS)
SINCE THE LONGITUDINAL STRESS DOMINATES, ASSUME A ONE DIMENSIONAL STRESS FIELD:
s s
x y
= =
s
Ee
z
x
=
0
THERE EXISTS A PLANE WHERE THE FIBERS DO NOT CHANGE LENGTH:
NEUTRAL AXIS
AT THE NEUTRAL AXIS:
ex = 0
SINCE THE BEAM IS BENT IN AN ARC, THERE EXISTS A CENTER OF CURVATURE AND A RADIUS OF CURVATURE - r
FROM CALCULUS:
1
r
d y 2 dx dy 1 dx
3/ 2 2
2
1
r
d2y dx2 dy 1 dx
2 3/ 2
IF THE DEFLECTION IS VERY SMALL:
dy dy 1 so 1 dx dx
2 3/ 2
1
d y 2 r dx
1
2
RECAP BASIC ASSUMPTIONS
THE BEAM LOCALLY BENDS INTO A CIRCULAR ARC. THE DEFLECTION IS SMALL. PLANE SECTIONS REMAIN PLANE. THERE IS A NEUTRAL AXIS WHERE THE LONGITUDINAL STRAIN IS ZERO. SHEAR DEFORMATIONS ARE IGNORED. ASSUME ONE-DIMENSIONAL STRESS.
Strain is linear sections (plane remain plane), y measured from neutral axis:
y e x e MAX c
LONGITUDINAL STRESS IS PREDOMINANT sy = sz = 0:
sx
=
Eex
ONE DIMENSIONAL HOOKE'S LAW
SINCE STRAIN IS LINEAR WITH RESPECT TO "y" AND STRESS IS LINEAR WITH RESPECT TO STRAIN, STRESS IS AS WELL LINEAR WITH RESPECT TO "y"
Stress is linear:
y s x s max c
FINDING THE NEUTRAL AXIS LOCATION
THE FORCE PERPENDICULAR TO A SURFACE IS "F":
F = dF = s dA
Equilibrium (if no external axial force):
Fx 0 s x dA
A
s max
c
y dA
E e max y dA c A
Since E, c and emax are constant:
Ee max Fx c
y dA 0
A
Since E, c and emax are non-zero:
A
y dA 0
Definition of centroid. Therefore neutral axis at centroid.
From a free-body diagram: External moment = Internal Moment
M s x ydA
M s x ydA
y s x s max c s max Mz y ( y )dA c s max 2 y dA c A
Mz
s max
c
y dA
2
A THE MOMENT OF INERTIA: (SECOND MOMENT OF AREA)
I z y dA
2
M z
s max I z
c
OR:
s max
M zc Iz
Stress is linear:
y s x s max c
Mzy s x Iz
STRESSES:
AXIAL
P s A
TORSION
Tr t J
My s I
BENDING
Example
A beam with the shown cross section is subjected to a positive moment of 50 k-in. Calculate the normal stress at the top, bottom, and interface between flange and web.
9" 1.5" 4.5" 3"
Example Cont.
Calculate the location of N.A.
9 " 1 " .5 4 " .5 3 " y
-
6 -y
-
Ay y A
i i
i
(9 1.5 1.5 / 2) 3 4.5 (1.5 4.5 / 2) 2.25" from the top (9 1.5) (3 4.5)
Example Cont.
Calculate the moment of inertia about the N.A.
9 " 1 " .5 4 " .5 3 " N . .A 2 5 .2 " 3 5 .7 "
1 I I Ad (9)(1.5)3 (9)(1.5)(2.25 - 1.5 / 2) 2 12 1 2 (3)(4.5) 3 (3)(4.5)(1.5 4.5 / 2) - 2.25 86.0625in 4 12
2
-
Example Cont.
Calculate the normal stresses
9 " s 1 1" . 5 4" . 5 N. . A 25 . " 2 35 . " 7
M0- . = kn 5 i
s 2 N. . S s 3
My s I 3 " 50 2.25 s1 1.3072 ksi 86.0625 50 (2.25 - 1.5) s2 0.4357 ksi 86.0625 50 3.75 s3 2.1786 ksi 86.0625
Se i w i V d e