For the loaded beam shown in the link below

1/For the loaded beam shown in the link below
2/Calculate the reactions at the supports
3/tell me how to draw the shear force and bending moment diagrams on paper
4/Find the position & magnitude of the maximum bending moment
(Relative to the left hand side)
And locate a point of contra flexure

http://public.blu.livefilestore.com/y1pFK7CcB6KIihzvqjQFiIwp1M2CXWevrL-2R7uYkVh5ZuH4d0_4kS0dbzFqCzwDaSY3Uk_3I-rhjpoY5Hk1SuYMQ/Untitled.pngn.png

Nomenclature:
Uniform load per unit length: w
Length: L
Left-most load: P
Right-most load: Q
Position of P (from left edge): a
Position of R2 (from left edge): b
Coordinate (from left edge): x

Your set of data is the following:
w:=22000 N/m; P:=40000 N; Q:=40000 N;
L:=7 m; a:=1 m; b:=6 m;

To find reactions at supports, first assess the uniform load as an equivalent concentrated load. Locate it at the center of the beam (@x = L/x), and its magnitude as w*L.

Do a moment balance about R1:
R2*b = w*L^2/2 + P*a + Q*L

Solve for R2:
R2 = (w*L^2/2 + P*a + Q*L)/b

Force balance to find R1:
R1 + R2 = w*L + P + Q

Solve for R1:
R1 = w*L + P + Q – R2

Or, with R2 substituted:
R1 = w*L + P + Q – (w*L^2/2 + P*a + Q*L)/b

When plugging in data, you get the following numbers:
R1 = 90833 Newtons
R2 = 143167 Newtons

———————————————————–
Observe the image, depicting the strategy of section cuts:

http://img710.imageshack.us/img710/4600/threesectbeam.gif

Split the beam in to regions. Draw a free body diagram of each region, and place unknown internal reactions of V and M at the cut-away location. By the standard sign convention, if the section is to the left, V is downward, M is counter clockwise.

For Region I:
Shear force, via force balance:
Vi = R1 – w*x

Bending moment via moment balance:
Mi = R1*x – 1/2*w*x^2

For Region II:
Shear force, via force balance:
Vii = R1 – w*x – P

Bending moment via moment balance:
Mii = R1*x – 1/2*w*x^2 – P*(x – a)

For Region III:
Shear force, via force balance:
Viii = R1 – w*x – P + R2

Bending moment via moment balance:
Miii = R1*x – 1/2*w*x^2 – P*(x-a) + R2*(x-b)

Plug in data, plot with your favorite plotting software. If you want to draw by hand, observe the results & patterns, and draw on your paper.

My plots are at the following image:

http://img710.imageshack.us/img710/4600/threesectbeam.gif

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To find the maximum bending moment, you can use the calculus optimization approach, or you can take a strengths shortcut. Shear is the derivative of bending moment, and therefore, when the shear equals zero, the bending moment is likely maximum.

We observe (according to the plots) both phenomena to occur in Region II.

Set the shear force to zero and solve for x.
Vii = 0, at a critical value of x
R1 – w*x – P = 0

Solve for x_crit:
x_crit = (R1 – P)/w

Plug in this result to expression for bending moment, region II:
After simplifying:
M_max = (R1^2 – 2*R1*P + P^2 + 2*P*a*w)/2*w

After plugging in data, we get our results:
x_crit = 2.310 meters
M_max = 98,728 Newton-meters

———————-
As for a point of contraflexure, this is a location where the bending moment equals zero, and the beam will have zero curvature. The obvious answers are x=0 and x=L, because they are at the end of the beam, and free to rotate.

But the not-so-obvious answer occurs in region II. Observe in the bending moment diagram.

Set Mii = 0, and solve for x:

Solving with a quadratic formula and accepting the only physical solution:
x = (R1-P+sqrt(R1^2-2*R1*P+P^2+2*w*P*a) )/w

Plug in data, and get result for contraflexure position:
x_contraflexure = 5.306 meters

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