We were
discussing shear force and bending moment
diagrams for a simply supported beam with a point load acting at midpoint of
the loaded beam and
shear force and bending moment diagrams for a simply supported beam with uniform distributed load during our previous posts.
We have
also seen shear force and bending moment
diagrams for a simply supported beam with an eccentric point load and shear force and bending moment diagrams for a simply supported beam with uniform varying load in our
recent post.
Today we
will see here the concept to draw shear force and bending moment diagrams for a
simply supported beam with uniform varying load from zero at one end to w per
unit length at the other end with the help of this post.
Let us see
the following figure, we have one beam AB of length L and beam is resting or supported
freely on the supports at its both ends. And therefore, displayed beam
indicates the simply supported beam AB and of length L.
Let us
consider that simply supported beam AB is loaded with uniformly varying load zero
at one end to w per unit length at the other end as displayed in following
figure.
First of
all we will remind here the important points for drawing shear force and
bending moment diagram. Now always remember that we will have to first
determine the reaction forces at each support.
Let us
consider that RA and RB are the reaction forces
at end support A and end support B respectively and we will use the concept of
equilibrium in order to determine the value of these reaction forces.
Before
going ahead, we must have to understand here the uniformly varying loading and
we can refer the post different types of load acting on
the beam in
order to understand the uniformly varying load.
In case of
uniformly varying load, total uniformly varying load will be converted in to
point load by determining the area of the triangle and will be acting through
the C.G of the triangle.
Total
uniform varying load = Area of the triangle ABO
Total
uniform varying load = (AB * BO)/2
Total
uniform varying load = w*L/2
ƩFX =0,
ƩFY =0, ƩM =0,
RA +
RB – w. L/2 =0
RA +
RB = w. L/2
ƩMB =0
RA *
L – w. L/2*(L/3) = 0
RA *
L = w. L2/6
RA =
w L/6
RB =
w L/3
Now we
have values of reaction forces at end A and end B and it is as mentioned above.
Let us determine now the value of shear force and bending moment at all
critical points.
Let us
consider one section XX between A and B at a distance x from end A. Now we will
have two portion of the beam AB i.e. left portion and right portion. Let us
deal with the left portion here. Let us assume that FX is
shear force at section XX and bending moment is MX at section
XX.
Shear force diagram
As we have
assumed here section XX between A and B at a distance x from end A and
therefore loaded beam AB will be divided in two portion and let us consider the
left portion of the beam. Shear force at section XX will be equivalent to the
resultant of forces acting on beam to the left side of the section.
Let us
first determine the rate of loading at section XX, because rate of loading at
section XX will be used during calculation of shear force.
Rate of
loading at section XX = XC in load diagram
XC/OB = x/
L
XC/w = x/ L
XC= w. x/L
Let us
determine the load of uniformly varying load up to section XX and we will have
Load of
uniformly varying load up to section XX = Area of triangle AXC
Load of
uniformly varying load up to section XX = (x. w. x/L)/2
Load of
uniformly varying load up to section XX = w.x2/2L
Load of
uniformly varying load up to section XX will act through a point
which will be at a distance of x/3 from section XX.
FX =
RA – w.x2/2L
FX =
w. L/6 – w.x2/2L
Force
acting to the left of the section and in upward direction will be considered as
positive and force acting to the left of the section but in downward direction
will considered as negative and we can refer the post sign conventions for shear force and
bending moment in
order to understand the sign of shear force which is determined here.
Now we
have information here in respect of shear force equation and therefore we will
determine shear force at all critical points i.e.
Shear
force at point A, x= 0
FA=
+ w. L/6
Shear
force at point B, x= L
FB =
w. L/6 – w.L2/2L = w. L/6 – w. L/2
FB=
- w. L/3
Let us
also determine the value of x for zero value of shear force, we can determine
as mentioned here
0 = w. L/6
– w.x2/2L
L2/3
= x2
Let us
consider that point C is at a distance of
Therefore
shear force will be zero at point C.
Again we
can refer the post sign conventions for shear force and
bending moment in
order to understand the sign of shear force which is determined here.
As we can
see that shear force is following parabolic equation and therefore we can now
draw here shear force diagram and it is displayed here in above figure.
Bending moment diagram
As we have
considered above one section XX between A and B at a distance x from
the end A and hence bending moment at section XX i.e. MX will
be determined as mentioned here.
MX =
RA. x – (w.x2/2L) x/3
MX =
(w. L/6).x – w. x3/6L
MX =
w. x. L/6 – w. x3/6L
As we know
that bending moment will be maximum at a point where shear force will be zero.
And as we have shown above that shear force will be zero at point C and hence
bending moment will be maximum at point C.
Bending
moment at point A, MA = 0
Bending
moment at point B, MB = 0
Please
refer the post Basics of shear force and bending
moment diagrams ,
in case of simply supported beam, bending moment will be zero at end supports.
Therefore
we have bending moment for all critical points and we have also information
about the equation followed by bending moment i.e. bending moment equation
stated above is following the cubic equation and therefore we can draw here the
BMD i.e. bending moment diagram and we have drawn it as displayed in above
figure.
Do you
have any suggestions? Please write in comment box
Reference:
Strength
of material, By R. K. Bansal
Image
Courtesy: Google
We will
see another important topic i.e. in the category of strength of material.
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