Question

Derive the 3-by-3 stiffness matrix for a beam element where the right end point is pin...

Derive the 3-by-3 stiffness matrix for a beam element where the right end point is pin supported such that the moment is zero (R4 = 0). The left end is fixed

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
overhanging beam is simply supported at san of 13m and carries of point load of 4000N...
overhanging beam is simply supported at san of 13m and carries of point load of 4000N 3m from right support the left overhangs is 3 m away from left support and has a point load of 5000N. the right overhang is of 4m carrying UDL of 200N per meter for entire rightover . draw shear force and bending moment diagram . calculate the maximum bending moment.
A beam is supported by two fulcrums and holds two masses, where m1= 50 g, m2...
A beam is supported by two fulcrums and holds two masses, where m1= 50 g, m2 = 550 g, the length of the beam is 150 cm, and the beam's mass is 210 g. The right fulcrum is 30 cm from the right end of the beam. Can m2 be placed on the beam so that there is no normal force at O due to the fulcrum on the left ? If not, record zero for the value of x....
using matlab: Q)using nested for loops to generate a 3*3 matrix where the element value is...
using matlab: Q)using nested for loops to generate a 3*3 matrix where the element value is equal to the sum of its row and column number, except for the diagonal elements which are zeros
Q3. Draw a SFD. BMD and the thrust diagram of a simply supported beam of span...
Q3. Draw a SFD. BMD and the thrust diagram of a simply supported beam of span 8m caring 3 point loads of 4kNand at 2m4mand each from left supportConsidering the right end to be degree these the loads are inclined at 150, 60 degrees ° and 45° respectively,
A simply-supported beam of length 13.3 inches has a distributed load applied along its entire length....
A simply-supported beam of length 13.3 inches has a distributed load applied along its entire length. The function defining the load is w(x) = 4x^2 + 13x - 6lb/ft with x measured in feet. 1) Determine the total applied load, in lb. 2) Determine the reaction at the left end of the beam (x = 0) 3) Determine the reaction at the right end of the beam (x = 13.3) 4) Determine and enter below the internal beam shear at...
Bordered Hessian element The Lagrangian is L=ln(x+y^2) -z^3/(3*y) -x*y +λ*(x*z +3*x^2*y -r), where r is a...
Bordered Hessian element The Lagrangian is L=ln(x+y^2) -z^3/(3*y) -x*y +λ*(x*z +3*x^2*y -r), where r is a parameter (a known real number). Here, ln denotes the natural logarithm, ^ power, * multiplication, / division, + addition, - subtraction. The border is at the top and left of the Hessian. The variables are ordered λ,x,y,z. Find the last element in the second row of the bordered Hessian at the point (λ,x,y,z) =(0.11, 0, 2440, 0.01167). This point need not be stationary and...
3. A uniform semicircle with a radius of 2.00 cm is placed, flat edge down, on...
3. A uniform semicircle with a radius of 2.00 cm is placed, flat edge down, on a horizontal removable platform. A pin is put in the left bottom corner to act as a pivot point and origin. (a) What is the x − y location of the center of mass? (b) What is the moment of inertia of the semicircle? (c) The platform is suddenly removed. Draw a free-body diagram of the semicircle, with axes and ± directions. (d) Using...
documentclass{article} \usepackage{array} \usepackage{tabulary} \usepackage{amsmath} \begin{document} C=capacitance of equivalent ckt.[7] \begin{equation} C=\dfrac{\epsilon_{ef}\epsilon_{o}L_{e}W}{2 h} F \end{equation} where\ \begin{center}...
documentclass{article} \usepackage{array} \usepackage{tabulary} \usepackage{amsmath} \begin{document} C=capacitance of equivalent ckt.[7] \begin{equation} C=\dfrac{\epsilon_{ef}\epsilon_{o}L_{e}W}{2 h} F \end{equation} where\ \begin{center} $F=\cos ^{ - 2} ({\pi}X_{f}/L)$ \end{center} L=inductance of equivalent ckt.[7] \begin{equation} L=\frac{1}{({2\pi}f_{r})^{2}C} \end{equation}\ $\Delta L$=additional series inductance \begin{equation} \Delta L=\frac{Z_{01}+Z_{02}}{16\pi{f_{r}}F} tan(\pi{f_{r}{L_{n}}}/C) \end{equation}\ $Z_{01} and Z_{02}$ are the characteristics impedances of microstrip lines with width of $w_{1} and w_{2}$ respectively.The values \begin{equation} Z_{01}=120\pi/(\frac{w_{1}}{h}+1.393+0.667\ln(\frac{w_{1}}{h}+1.444)) \end{equation}\ $$ Z_{02}=120\pi/(\frac{w_{2}}{h}+1.393+0.667\ln(\frac{w_{2}}{h}+1.444)) $$ \ where\ $$w_{1}=w-2{P_{s}}-W_{s}$$ \ and\ $$w_{2}=2{P_{s}}-W_{s}$$ The capacitance$\Delta C$ between center wing and side wing is calculated as...
Two point charges are fixed in place along the x axis as shown in the figure...
Two point charges are fixed in place along the x axis as shown in the figure above. The charge Q1=-3 μμC is located at the origin. The charge Q2=4 μμC is a distance a = 0.46 m to the right of the origin. 1) Calculate the x component of the electric field at point P a distance h=0.32 m above the origin. Ex = -0.115 × 106 N/C Ex = -0.094 × 106 N/C Ex = 0.264 × 106 N/C...
1) 2 point charges are separated by a distance of 8 cm. The left charge is...
1) 2 point charges are separated by a distance of 8 cm. The left charge is 48 mC and the right charge is -16mC. Using a full sheet of paper: draw the 2 charges separated by 8cm, centered in the sheet. (if you are missing a ruler estimate 8cm as ⅓ a paper sheet length). [6] a) Draw field lines to indicate the electric fields for this distribution. [4] b) Draw 3 equipotential surfaces, 1 each, that pass: -Through the...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT