Last Updated on February 12, 2024 by Maged kamel

List of compression Member Posts-part 3.

Solved problem 4-16 using the French equation.

This is the 21st post of the compression Member Posts-part 3, which includes our problem is 4-16, from Prof. Alan Williams’s book, structural engineering reference manual. The French equation was made For side sway Uninhibited- unbraced frame, for the k value, which is =sqrt(1.6*GA*Gb+4(Ga+Gb)+7.5)/ sqrt of(GA+Gb+7.5).

This is the list of post 21: Solved problem 4-16 using the French equation.

Modification to Alignment Chart Braced Frame.

This is the 22nd post of the compression Member Posts-part 3 which includes how to consider is the adjustment of columns of different end conditions ? and how to get the m value for end conditions for girders of braced frames?

This is the list of post 22: Modification to Alignment Chart Braced Frame.

Solved problem 7-2 for frames-Two parts a,b.

This is the 23rd post of the compression Member Posts-part 3 which includes Example 7.2. from Prof. Mccormack’s book. Determine the k values determined and shown in the figure K factors for each of the columns of the frame shown in Fig. 7.6.

Here, W sections have been tentatively selected for each of the members of the frame, and their I/L values are determined and shown in the figure.

This is the list of post 23: Solved problem 7-2 for k for frame 1/3.

This is the list of post-23a: Solved problem 7-2 for frames

Solved problem 7-2 for frames-part 3.

This is the 24th post of the compression Member Posts-part 3 which includes starting by joint H, and joint H at joint H no upper column, G=sum(EI/L) which is=20.47/ sum(EI/L) for girder Gh =26.67, Gh=(20.47/26.67)=0.675.

For Gg at joint G, we have two columns GH, GF, Gc=(20.47+31.67)/sum of(mEI/L) for girders CG and GJ, m value=1 at GC and m= 1.5, for girder GJ because of the hinge at far joint, the denominator =(1*70+1.5*21.25), Gg=0.5118.

This is the list of post-24: Solved problem 7-2 for frames-part 3.

Stiffness reduction factor for inelastic columns.

This is the 25th post of the compression Member Posts-part 3 which includes the difference between the modulus of elasticity E and the other modulus of elasticity Et, which is the tangent modulus of elasticity, and in what conditions for which we will use the Et?

The second point, after God’s well, τb, or the stress reduction factor, is for the case of short or medium columns.

This is the list of post-25: Stiffness reduction factor for inelastic columns.

Solved problem 4-13 for the stiffness reduction factor.

This is the 26th post of the compression Member Posts-part 3 which includes Example 4.13, from Prof. Alan Williams’s book. A W10 ×54 of A992 steel is used as a column. It is subjected to a service dead load of 100 kips and a service live load of 200 kips.

If the slenderness ratio makes this member an inelastic column, what is the stiffness reduction factor, tb?

This is the list of post-26: Solved problem 4-13 for the stiffness reduction factor.

Solved problem 4-14 for the effective length factor.

This is the 27th post of the compression Member Posts-part 3 which includes Example 4.14, from Prof. Alan Williams’s book. A rigid unbraced frame is shown in Figure 4.17. All members are oriented so that bending is about the strong axis.

Lateral support is provided at each joint by simply connecting bracing in the direction perpendicular to the frame. Determine the effective length factors with respect to each axis for member AB. The service dead load is 35.5 kips, and the service live load is 142 kips. A992 steel is used.

This is the list of post-27:Solved problem 4-14 part-1 for effective length factor.

This is the second link to post 28: Solved problem 4-14 part 2 for effective length factor.

Here is the list of compression members in part-2.

An external source for Compression members from Prof. T. Bartlett Quimby’s site.