Discussion on: Solution to Problem 535 | Economic Sections
Following is a discussion on the Reviewer item titled: Solution to Problem 535 | Economic Sections. Feel free to add your own comments!
- Log in to post comments
Active forum topics
- Deriving trigonometric function
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
- Application of Differential Equation: Newton's Law of Cooling
- Minima maxima: a²y = x⁴
New forum topics
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
- Required diameter of solid shaft
- Deriving trigonometric function
- Integration $\displaystyle \int \sec 2x ~ dx$
535 ECONOMIC SECTION
Mr. Verterra,
What is the rationale behind calling y=124 lbs. 14 ft. from R2 a critical section? You will see this in computing for the maximum moment of the dead load. I just need to remember why?
Thank you,
Benjamin A. Molina III
Re: 535 ECONOMIC SECTION
In reply to 535 ECONOMIC SECTION by Benjamin_Molina
I am not Mr. Verterra but from the live load diagram, it is the location of zero shear.
Re: Discussion on: Solution to Problem 535 | Economic Sections
The critical section for moment will usually occur at the point of zero shear. From the shear diagram of the applied loads, the point of zero shear is located 14 meters from the right support. The maximum moment will occur at that point.
Considering the effect of the weight of the member (which is significantly small compared to the applied load), the location of maximum moment won't change (if it does, it is only very small - the solution here did not check if it does). Thus, the critical point for bending is still 14 meters from the right support.