SFD and BMD - Problem 1 - Part 1 - Shear Force and Bending Moment Diagram - Strength of Materials
Summary
TLDRThis script details the process of analyzing an overhanging beam's structural behavior under various point loads. It begins with calculating support reactions at points A and B using equilibrium conditions, resulting in RA = 16.71 kN and RB = 32.09 kN. The script then proceeds to determine the shear force diagram (SFD) and bending moment diagram (BMD), crucial for understanding the beam's stress distribution. The SFD reveals varying forces along the beam, while the BMD is essential for identifying the point of contraflexure, a key structural feature. The explanation is technical, targeting an audience with an engineering background.
Takeaways
- đ The task involves analyzing a structural engineering problem concerning an overhanging beam with specific support points and loads.
- đ The primary goal is to determine the support reactions at points A and B, which are essential for understanding the beam's stability.
- âïž The equilibrium condition is applied by summing vertical forces (FY) to zero, considering upward forces as positive and downward forces as negative.
- đ Moments are calculated about point A, with clockwise moments being positive and anti-clockwise moments negative, to find the reaction force RB.
- đą Through calculations, the reaction force RB is determined to be 32.29 kilo Newtons, which is a critical value for the beam's support.
- đ The reaction force RA is then found by subtracting RB from the total vertical force, resulting in RA being 16.71 kilo Newtons.
- đ Shear force diagrams (SFD) are to be drawn to visualize how the forces vary along the length of the beam.
- đ The sign convention for shear force calculations is established, with upward forces to the left and downward forces to the right considered positive.
- đ Shear force at point C is calculated considering the forces and reactions, resulting in a value that remains constant up to point A.
- đ The shear force changes at different sections, such as at point D and B, reflecting the impact of the loads and reactions on the beam's integrity.
- đ The final step involves identifying the point of contraflexure, which is a location along the beam where the curvature changes sign, indicating a change in the bending moment.
Q & A
What is the main objective of the script?
-The main objective of the script is to solve a structural engineering problem involving an overhanging beam, which includes determining support reactions at points A and B, drawing the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and finding the point of contraflexure.
What are the given point loads on the overhanging beam?
-The given point loads on the overhanging beam are 10 kN at 2 meters from A, 24 kN at 4 meters from A, and 15 kN at 10 meters from A.
What are the equilibrium conditions used to find the support reactions?
-The equilibrium conditions used are the summation of vertical forces (ÎŁFY) and moments about a point (ÎŁMA), which are set to zero according to the principles of static equilibrium.
What is the calculated reaction at support A (RA)?
-The calculated reaction at support A (RA) is 16.71 kN.
What is the calculated reaction at support B (RB)?
-The calculated reaction at support B (RB) is 32.29 kN.
How are the moments about point A calculated?
-The moments about point A are calculated by considering the clockwise moments as positive and the anti-clockwise moments as negative, and summing these moments to zero.
What is the sign convention for shear force calculations?
-The sign convention for shear force calculations is that upward forces to the left and downward forces to the right are positive, while downward forces to the left and upward forces to the right are negative.
What is the shear force at point C?
-The shear force at point C is -10 kN, considering the downward force to the left of the section as negative.
How does the shear force change from point A to point D?
-The shear force at point A is +6.71 kN, and it remains constant up to the left of point D. To the right of point D, the shear force becomes -17.29 kN after considering the reactions and loads.
What is the shear force at point B?
-The shear force at point B is +15 kN, which remains constant up to point E.
What is the next step after calculating the support reactions and shear forces?
-The next step after calculating the support reactions and shear forces is to draw the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and then find the point of contraflexure.
Outlines
đïž Structural Analysis of Overhanging Beam
The first paragraph introduces a structural engineering problem involving an overhanging beam with specific support points and point loads. The task is to determine the support reactions at points A and B, draw the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and identify the point of contraflexure. The solution begins with the equilibrium condition, calculating the vertical forces RA and RB, treating upward forces as positive and downward forces as negative. The calculation leads to an equation involving RA and RB, which are then solved using moment equilibrium around point A, yielding RA as 16.71 kN and RB as 32.29 kN. This section concludes with the determination of the support reactions.
đ Shear Force Calculations for Beam Structure
The second paragraph continues the structural analysis by detailing the process of calculating the shear forces along the beam. The sign convention for shear force is established, with upward forces to the left and downward forces to the right considered positive, and vice versa. The calculation starts by examining the section close to point C, determining the shear force at various points along the beam, including at points A, D, B, and E. The shear force values are calculated by considering the forces acting to the left and right of a chosen section, leading to a constant shear force between certain points. The final values determined are -6.71 kN at point A, -17.29 kN at point D, and +15 kN at points B and E, concluding the shear force calculations for the beam.
Mindmap
Keywords
đĄOverhanging Beam
đĄSupport Reactions
đĄEquilibrium Condition
đĄShear Force Diagram (SFD)
đĄBending Moment Diagram (BMD)
đĄPoint of Contraflexure
đĄKilo Newton
đĄMoments
đĄSign Convention
đĄShear Force
đĄBending Moment
Highlights
The task is to determine support reactions at points A and B for an overhanging beam.
The beam has point loads of 10, 24, and 15 kilo Newtons.
Equilibrium conditions are applied to calculate support reactions RA and RB.
Vertical forces are considered with upward forces as positive and downward as negative.
Summation of vertical forces (ÎŁFY) is set to zero for equilibrium.
Moments about point A are calculated with clockwise moments as positive and anti-clockwise as negative.
Moment calculations involve distances from point A and the direction of forces.
RA and RB values are determined using equilibrium equations.
RA is calculated to be 16.71 kilo Newtons.
RB is determined to be 32.29 kilo Newtons.
Shear force diagram (SFD) and bending moment diagram (BMD) are to be drawn next.
Sign convention for SFD is established with upward forces to the left as positive.
Shear force at point C is calculated considering the forces to the left of the section.
Shear force at point A is determined by considering the forces on the left of section A.
Shear force at point B is calculated in two ways, yielding the same result of +15 kilo Newtons.
Shear force at point E is constant and equal to +15 kilo Newtons.
The process involves calculating the point of contraflexure, though not detailed in the transcript.
Transcripts
Let us take this first question for an
overhanging beam shown in Figure
determine support reactions at A and B
draw SFD in BMD and find the point of
contraflexure now this is the question
which we have in front of us let us try
to write the data first here I will draw
the beam which is given
now this is the question which we have
in front of us where we have an
overhanging beam which is supported at
Point A at point B then we have point
load of 10 kilo Newton 2400 10 and 15
kilo Newton acting on this beam for this
diagram we have to find out the support
reactions the first question is we have
to calculate reaction at a and reaction
at B that is I have to determine our a
value and our B value the next part is
we have to calculate s FD and bmd and
after that we have to find the point of
contraflexure so this is the question
which we have I will write the solution
for this I'll say that for the
equilibrium condition of the beam that
is my step number one is calculation of
support reactions here I will say that
for the equilibrium condition first
summation of FY is equal to zero where I
will treat upward forces positive and
downward force is negative so therefore
I have the sum of vertical forces as RA
is upward RB is upward so both are
positive that is RA plus RB 10 24 and 15
they are downwards so negative minus 10
minus 24 and minus 15 is equal to 0
therefore we have RA plus RB is equal to
if I add these terms it will be 10 plus
24 plus 15 which which is 49 here I have
negative 49 if I shift it on to the
other sides it becomes positive
so here I have my first equation after
this I will be taking moments at Point a
so I will say that summation of moments
about point a is equal to 0 where
clockwise moments are positive and
anti-clockwise moments are negative so
therefore if I am taking moment at Point
a I start from R be the moment is
anti-clockwise and distance is 7 meter
between point a and B so it is minus RB
into 7 next here I have 15 kilo Newton
which is at a distance of 10 meters from
point A and it is in clockwise direction
so I will take it as plus 15 into 10
then I have 24 kilo Newton which is at 4
meters from a and it is in clockwise
direction so plus 24 into 4 next I have
10 its distance is 2 meters from point A
and if I take the moment about point a
it is an anti-clockwise direction so
minus 10 into 2 therefore if I calculate
all terms here I will be getting RB as
it comes out to be 32 point 2 9 kilo
Newton and after this I will say that
put RB is equal to 32 point 2 9 kilo
Newton
in equation 1 so therefore I have RA it
was plus RB so plus 32 point 2 9 is
equal to 49
therefore RA is equal to 49 minus 32
point two nine
therefore RA value it comes out to be
sixteen point seven one kilo Newton if I
subtract these values so here I have
found out the support reactions and this
completes the first part that is I have
got the first answer RA RB values now I
will be writing the values here
ra is 16.7 1 RB is 32 point O nine so
here I have found the support reactions
the next part is I have to calculate
shear force so my step number two is SF
calculations and for that I will draw
the sign convention first upward force
to left and downward force to the right
are positive downward force to left and
upward force to right they are negative
so therefore I will start calculating
the value of shear force first I will be
taking the section at you can say very
close to Point C this is my section I
will see that first SF at Point C
is equal to here I have a section and to
the left of it that is downward force so
downward force to the left is taken as
negative value is minus 10 kilo Newton
next this shear force will remain
constant between C to a up to left of a
now when I take the section to the right
of a and if I look to the left of this
section I have minus 10 and plus sixteen
point seven one so SF at Point a minus
10 plus sixteen point seven one that
comes out to be plus six point seven one
kilo Newton now this value of shear
force at Point a will remain constant
from A to D up to the left of D if I
take the section to the right of D then
I have minus ten plus sixteen point
seven one minus 24 this value of shear
force it comes out to be minus seventeen
point two nine kilo Newton next SF at
point B is equal to if I take the
section to the right of B here I can
calculate it in two different ways
either I can see the left of section or
right of the section the answer will be
same so here if I take left I will have
minus ten plus sixteen point seven one
up hold
- 24 downward +32 0.09 upward so I have
the value as plus 15 kilo Newton and
this value will remain constant up to
point E so I say that SF at Point E is
equal to plus 15 kilo Newton so these
are the SF calculations
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