The post includes the product of inertia Ixy for the triangular section case -2. Two alternative methods were used. A full description is introduced.
12-Product of inertia Ixy for the right angle-case-2. Read More »
In our post, we continue solving the Solved problem-5-1-part 2-2 in which the design of steel beam is governed by deflection.
24b- Easy approach to solved problem-5-1-part 2-2. Read More »
TIn our post we will introduce a solved problem-5-1-part 1-2 in which the design of steel beam is governed by deflection.
24a- Easy approach to the solved problem-5-1-part 1-2. Read More »
Step-by-step guide on how to derive the expression for L-U values for-3×3 matrix? Matrix A is converted to U1 and L1 matrix, then to L2 and U2 matrix.
4-Derive the expression for L-U values for-3×3 matrix. Read More »
A review of portal frames subject, how to derive the internal forces for the portal frame with hinged supports at the base?
18-A Review of portal frames-pin supports at base. Read More »
How to get the moment of inertia for a right-angle triangle -Iy-case-2?The procedure is done by using a vertical strip for integration. how to find polar moment of inertia Jo?
11-Moment of inertia Iy-Case-2 for a right-angle triangle. Read More »
Deflection of steel beams, what are the disadvantages. What are the parameters controlling the deflection of steel beams?
24- Easy approach to deflection of steel beams. Read More »
How to get Ix-the moment of inertia for the right-angle triangle-Case-2? by using a horizontal strip, and the other method is to use a vertical strip.
10-Ix-the moment of inertia for right-angle-Case-2. Read More »
How to solve for x-y by L-U Decomposition? Illustration by Gauss elimination for L and U matrices.
3-How to solve for x-y by L-U Decomposition? Read More »
Alignment chart-part 2, discussion about the conditions from which the alignment chart was developed and an introduction to the portal frame.
17-Alignment chart-part 2 for the unbraced frames. Read More »
