Influence Line (02)

Dear Viewers,
Some days ago I have presented to you an article about "Influence Line", I hope all of you can remembered that,

http://seasoft022.blogspot.com/2013/05/influence-line.html

According to that article I am going to present to you another article about Influence Line, which is Influence Line (02). In this article I will tell you more information/rules about Influence Line with example.
So, viewers, Lets Start,

At first,
Necessary Information about Influence Line:

01.      Influence Line will always in Straight Line, not in Curved Line.
02.      Influence Line will not change its own slope over the related section.
03.      Shear at a point: Shear at a point means to cut at that point.
            
            Shear Positive when,
            ∑F_y=(summation of vertical forces) = Upward at Left section,
            ∑F_y=(summation of vertical forces) = Downward at Right section, and

            Shear Negative when,
            ∑F_y=(summation of vertical forces) = Downward at Left section,
            ∑F_y=(summation of vertical forces) = Upward at Right section
   
In case of Shear Calculation,
when 1 kip load (1 kip load is the standard load value for Influence Line) acts at the left side of Shear Point then we will calculate the reaction values at right side of Shear Point and,

when 1 kip load (1 kip load is the standard load value for Influence Line) acts at the right side of Shear Point then we will calculate the reaction values at left side of Shear Point.

Now Lets See an example to understand the term "Shear" very clearly,

Assume, AB is a beam of 10 feet, now we will apply Shear at point C (middle point of AB, so AC=BC=5 feet),
Now at first when 1 kip vertically downward load acts at Left Side which moves from A to C_L (C_L=Just immediate Left of C), then Shear at C will be,
V_C=Shear at Point C= (- R_B), because Shear will be Negative (following picture-01) when

 ∑F_y=(summation of vertical forces) = Upward at Right side,

Picture-01: When 1 kip load acts at the left portion of Shear Point C

Correspondingly, when 1 kip vertically downward load acts at Right Side which moves from C_R to B, (C_R=Just immediate Right of C), then Shear at C will be,
V_C=Shear at Point C= (+ R_A), because Shear will be Positive (following Picture-02) when

 ∑F_y=(summation of vertical forces) = Upward at Left Side,

Picture-02: When 1 kip load acts at the right portion of Shear Point C

Analysis of a Simply Supported Beam:

Consider the Following Simply Supported Beam,
AB= 10 feet, C is the middle point of AB, So that AC=CB=5 feet,

Picture-03: AB is a Simply Supported Beam where AB= 10 feet, C is the middle point of AB, So that AC=CB=5 feet

Now, we will draw the Influence Line (IL) for  R_A, Influence Line (IL) for R_B , Influence Line (IL) for V_C (Shear at Point C), Influence Line (IL) for M_C (Moment at Point C).

Influence Line (IL) for R_A:

Picture-04: Influence Line (IL) for R_A for above Simply Supported Beam AB

Influence Line (IL) for R_B:

Picture-05: Influence Line (IL) for R_B for above Simply Supported Beam AB

Influence Line (IL) for V_C:

Picture-06: Influence Line (IL) for V_C for above Simply Supported Beam AB

Influence Line (IL) for M_C:

Picture-07: Influence Line (IL) for M_C for above Simply Supported Beam AB

For any simply supported beam or, any beam with simply supported portion, we can draw the Influence Line (IL) by the above formulas/rules.



Now we are going to draw Influence Line for a Beam with mathematically calculation,
Let's Start,

Example-01:

Picture-08: Question for Example-01

Draw the Influence Line for R_A , R_C , V_B , M_B.

Solution:

_{When 1 kip load acts at point A, then we get,}

Picture-09: When 1 kip load acts at point A

When 1 kip load acts at point B, then we get,

Picture-10: When 1 kip load acts at point B

When 1 kip load acts at point C, then we get,

Picture-11: When 1 kip load acts at point C

When 1 kip load acts at point D, then we get,

Picture-12: When 1 kip load acts at point D

Now, Influence Line (IL) for R_A:
When 1 kip load at point A then R_A= 1,
When 1 kip load at point B then R_A= 0.5,
When 1 kip load at point C then R_A= 0,
When 1 kip load at point D then R_A= -0.6,
So, we will get Influence Line (IL) for R_A which is as follows,

Picturre-13: Influence Line (IL) for R_A

Now, Influence Line (IL) for R_C:
When 1 kip load at point A then R_C= 0,
When 1 kip load at point B then R_C= 0.5,
When 1 kip load at point C then R_C= 1,
When 1 kip load at point D then R_C= 1.6,
So, we will get Influence Line (IL) for R_C which is as follows,

Picture-14: Influence Line (IL) for R_C

Now, we will go calculation for shear,

When 1 kip load moves from A to B_L then, V_B= -R_C, (Right Side upward so, negative),
When 1 kip load moves from B_R to D then, V_B= +R_A, (Left Side upward so, positive),
When 1 kip load at point B, then , V_B= -R_C, (when just immediate left at B) and
                                                         V_B= +R_A, (when just immediate right at B) both of these, at first V_B= -R_C, then V_B= +R_A, because 1 kip load moves from Left to Right,
So, value for Left Side will be the first.

Now, Influence Line (IL) for V_B:
When 1 kip load at point A then V_B= -R_C, So, V_B=0, [when 1 kip load at point A, R_C= 0]
When 1 kip load at point B (B_L) then V_B= -R_C, So, V_B=-0.5, [when 1 kip load at point B, R_C=0.5]
When 1 kip load at point B (B_R) then V_B=+R_A, So, V_B=+0.5,[when 1 kip load at point B, R_A=0.5] 
When 1 kip load at point C then V_B=+R_A, So, V_B=0, [when 1 kip load at point C, R_A= 0]
When 1 kip load at point D then V_B=+R_A, So, V_B=-0.6, [when 1 kip load at point D, R_A=-0.6]

So, we will get Influence Line (IL) for V_B which is as follows,

Picture-15: Influence Line (IL) for V_B

Though, 1 kip load moves from left to right then in case of point B, V_B is considered (-0.5) at first then we have considered (+0.5).


Now, we will go calculation for Moment,
When 1 kip load moves from A to B_L then, M_B=R_C*5 
When 1 kip load moves from B_R to D then, M_B=R_A*5 
When 1 kip load at point B, then , M_B= (R_C*5) or, (R_A*5), in both cases the value of M_B will be the same.

Now, Influence Line (IL) for M_B:
When 1 kip load at point A then M_B= (R_C*5), So, M_B=0, [when 1 kip load at point A, R_C= 0]
When 1 kip load at point B then M_B= (R_C*5) or (R_A*5) So, M_B= 0.5*5=2.5, [when 1 kip load at point B, R_A=R_C=0.5]
When 1 kip load at point C then M_B=R_A*5, So, M_B=0, [when 1 kip load at point C, R_A= 0]
When 1 kip load at point D then M_B=R_A*5 So, M_B=(-0.6*3), M_B=3, [when 1 kip load at point D, R_A=-0.6]

So, we will get Influence Line (IL) for M_B which is as follows,

Picture-16: Influence Line (IL) for M_B

Now if we want to create a Data Chart by collecting these values, then we get the following chart,

Picture-17: Data Chart for Example-01

Dear Viewers,
May be this article as well as this example become very large, but it is not large actually, Because of the necessary explanation, the solution of this mathematical problem become very lengthy/long. But if you can understand "How to draw Influence Line (IL)" clearly then from the next time this lengthy explanation will not be needed, so we could draw Influence Line easily in future. Thank You.

Thanks and Regards,
Engr. Snehashish Bhattacharjee (Tushar),
SEA Soft and Design Consultants, seasoft022.blogspot.com

Shear Force & Bending Moment Diagram (03)

Dear Viewers,

With reference to my previous two (02) articles about Shear Force & Bending Moment Diagram, I am going to present my another article about Shear Force & Bending Moment Diagram. At first let review the previous articles once again.

My first article contains the basics rules of Shear Force Diagram (SFD) & Bending Moment Diagram (BMD), which is,

http://seasoft022.blogspot.com/2013/07/shear-force-bending-moment-diagram-01.html

My second article contains the 5 (Five) examples about Shear Force Diagram (SFD) & Bending Moment Diagram (BMD), which is,

http://seasoft022.blogspot.com/2013/07/shear-force-bending-moment-diagram-02_4.html

Now, I am going to present to you more 3 (three) examples about Shear Force Diagram (SFD) & Bending Moment Diagram (BMD), So Let's Start,

Example-01:

A beam (ABCD, AB=3 feet, BC=3 feet, CD=2 feet) of total 8 feet length, Pin Support acts at point A which is most left point of this beam, Roller Support acts at point D which is most right point of this beam. UDL 1 kip/feet acts over AB portion (3 feet) and 3 kip/feet triangular UDL acts over CD portion (2 feet).

Example-01: Page 01 of 02

Example-01: Page 02 of 02

Example-02:

A beam (ABCDEFG, AB=15 feet, BC=15 feet, CD=5 feet, DE=5 feet, EF=10 feet, FG=5 feet) of total 55 feet length, Pin Support acts at point B which is 15 feet far from left end, Roller Support acts at point D which is 20 feet far from right end. UDL 1 kip/feet acts over EF portion (10 feet) and 2 kip/feet triangular UDL acts over AB portion (15 feet). Angula Point Load 50 kip acts at point C where horizontal:vertical ratio is 3:4.

Example-02: Page 01 of 02

Example-02: Page 02 of 02

Example-03:

A beam (ABCD, AB=6 feet, BC=4 feet, CD=3 feet) of total 13 feet length, Triangular UDL 2 kip/feet acts over AB portion (46feet) and Triangular UDL 4 kip/feet triangular UDL acts over CD portion (3 feet). An UDL of 3 kip/feet acts at the portion BC (4 feet) with upward direction.

Example-03: Page 01 of 01

Dear Viewers,
If my articles about Shear Force Diagram (SFD) & Bending Moment Diagram (BMD) help you, that will my success and my hard-working will get its proper award. If you have any more question about SFD & BMD then please write me. I will try my best to give answer your question. Please feel free to contact for any other information.

Thanks and Regards,
Engr. Snehashish Bhattacharjee (Tushar),
SEA Soft and Design Consultants, seasoft022.blogspot.com
E-mail: sea.bd.022@gmail.com