Discussion- Simplified Design for Torsional Loading of Rolled Steel Members (1)

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Discussion Simplified Design for Torsional Loading of Rolled Steel Members Paper presented by PHILIP H. LIN (3rd Quarter, 1977) Discussion by Nirmal Maitra The author has done a good service by presenting tables to accelerate design work for beams subjected to torsional moment. The long calculation to convert torsional moment to equivalent lateral moment for obtaining torsional bending stress can very w^ell be avoided by using the au- thor's tables. The author has, how^ever, neglected one very important element, i.e., torsional shear stress. The reason, he states: "In most practical steel design situations, the shear stress contributions are normally not of significance; therefore, only the normal stresses from vertical and torsional bending are computed and combined in order to check the design adequacy of a beam." The writer does not agree with him. As a case for checking, take author's Example 1. The author's calcu- lation shows that summation of bending stresses does not exceed 0.6Fy and he concludes ''use WIO X 54". But the calculation presented below shows that summation of shear stresses exceeds the allowable limit by 60% and WIO X 54 is no good. Equipment supporting structures usually have the tor- sional loading concentrated at points similar to what is shown in author's Examples 1, 2, and 4, and in such cases shear stresses usually govern. A check on the author's examples will prove that. A prevalent design office practice is to check torsional shear stress before taking up calculations for torsional bending stress for equipment supporting beams; the merit of this practice is obvious. Calculation: For WIO X 54: t^/tf= 0.368/0.618 = 0.6 R/tf= 0.5/0.618 = 0.8 Nirmal Maitra is Senior Structural Engineer, American Lurgi Corporation, Hasbrouck Heights, N.J. From chart,* 6 = 1.25 Neglecting warping restraint: f^T = ^Ttf/J = 1.25 X 4.5 X 12 X 0.618/1.84 = 22.67 ksi U = F/^z^ = 9/(10.12X0.368) = 2.42 ksi Summation: 22.67 + 2.42 = 25.09 ksi >14.5 ksi n.g. Try WIO X 77 and repeat checking in above manner. Found o.k. Use: WIO X 77 26

description

Design for Torsion as per AISC

Transcript of Discussion- Simplified Design for Torsional Loading of Rolled Steel Members (1)

Page 1: Discussion- Simplified Design for Torsional Loading of Rolled Steel Members (1)

Discussion

Simplified Design for Torsional Loading of Rolled Steel Members

Paper presented by PHILIP H. LIN (3rd Quarter, 1977)

Discussion by Ni rmal Mai t ra

The author has done a good service by presenting tables to accelerate design work for beams subjected to torsional moment. The long calculation to convert torsional moment to equivalent lateral moment for obtaining torsional bending stress can very w^ell be avoided by using the au­thor's tables.

The author has, how^ever, neglected one very important element, i.e., torsional shear stress. The reason, he states: " In most practical steel design situations, the shear stress contributions are normally not of significance; therefore, only the normal stresses from vertical and torsional bending are computed and combined in order to check the design adequacy of a beam."

The writer does not agree with him. As a case for checking, take author's Example 1. The author's calcu­lation shows that summation of bending stresses does not exceed 0.6Fy and he concludes ' 'use WIO X 54". But the calculation presented below shows that summation of shear stresses exceeds the allowable limit by 60% and WIO X 54 is no good.

Equipment supporting structures usually have the tor­sional loading concentrated at points similar to what is shown in author's Examples 1, 2, and 4, and in such cases shear stresses usually govern. A check on the author's examples will prove that.

A prevalent design office practice is to check torsional shear stress before taking up calculations for torsional bending stress for equipment supporting beams; the merit of this practice is obvious.

Calculation:

For WIO X 54: t^/tf= 0.368/0.618 = 0.6

R/tf= 0.5/0.618 = 0.8

Nirmal Maitra is Senior Structural Engineer, American Lurgi Corporation, Hasbrouck Heights, N.J.

From chart,* 6 = 1.25

Neglecting warping restraint:

f^T = ^Ttf/J = 1.25 X 4.5 X 12 X 0.618/1.84

= 22.67 ksi

U = F/^z^ = 9 / (10 .12X0 .368)

= 2.42 ksi

Summation:

22.67 + 2.42 = 25.09 ksi >14.5 ksi n.g.

Try WIO X 77 and repeat checking in above manner.

Found o.k.

Use: WIO X 77

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Page 2: Discussion- Simplified Design for Torsional Loading of Rolled Steel Members (1)

Simplified Design for Torsional Loading of Rolled Steel Members

Paper by PHILIP H. LIN (3rd Quarter 1977)

Discussion by Philip H. Lin

There are two types of torsion, namely, uniform and non­uniform torsion. A straight round bar will rotate freely when subjected to two equal and opposite torques at the ends, and the twist per unit length or the angle of twist is constant throughout the bar; this is known as uniform torsion. When a member is under certain end boundary conditions, however, and torques somewhere along its length, it will be restrained to warp and the angle of twist

Philip H. Lin is Senior Structural Engineer, The Ralph M. Parsons Company, Pasadena, Calif.

will vary; this is known as non-uniform torsion. Because of the difference in the structural behavior of these two types of torsion, the treatment required is obviously not the same for both.

Examples presented in the original paper are cases of non-uniform torsion, problems encountered frequently in the design office. In Example 1, the differential equation solution yields the results shown in Table 1. By investi­gation of the stresses in Table 1, one can comfortably design the beam based on normal stress. One can also see that the the summation of the shear stresses is very low, and thus not a governing factor in this structural steel design situa­tion.

In his discussion of the subject paper (1st Quarter, 1978), Mr. Maitra used Eq. (7.17) in his quoted reference to check the shear stress of Example 1 and concluded that the beam is 60% overstressed. This, I believe, is not the case, because Eq. (7.17) is derived based on the concept of uniform tor­sion. It is not adequate to apply the equation for uniform torsion to Example 1, which deals with non-uniform tor­sion. Although it is left to the design engineer's discretion to use that equation to check torsional stress before going into design, it is not my recommendation, for one should always bear in mind that the structural member so designed will be extremely conservative. There are exceptions, however, e.g., the design of a highway sign post, where it is very likely the shear stress would govern the design. If that is the case, the torsional shear stress should be checked. (The closed shape structural section is recommended.)

As shown in Table 1, the total normal stress is the summation of torsional and ordinary bending stresses, and the total shear stress at flange or at web is the summation of St. Venant torsion, warping torsion, and vertical bending shear stresses as shown in Fig. 6 of the original paper.

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ENGINEERING JOURNAL/AMERICAN INSTITUTE OF STEEL CONSTRUCTION