Question 6.9: The beam in Fig. 1, an S12 50 I-beam, is subjected to a mome......
The beam in Fig. 1, an S12×50 I-beam, is subjected to a moment M = 150 kip · in. that is supposed to lie along the z axis (e.g., due to loading in the xy plane). (a) Determine the effect of a “load misalignment” of θ = 2° on the orientation of the neutral axis. Show the neutral axis on a sketch of the cross section; and (b) determine the maximum tensile stress and the maximum compressive stress on the section. Note the large difference in I_y and I_z, and note how this affects the solution.
Plan the Solution We can use Eq. 6.33 (or Eq. 6.31) to determine the orientation of the neutral axis. Then we can identify the two points farthest from the NA and use Eq. 6.30 to compute the flexural stress at these points.
\tan β = \left(\frac{I_z}{I_y}\right) \tan θ (6.33)
\left(\frac{M_y}{I_y}\right)z^∗ – \left( \frac{M_z}{I_z}\right)y^∗ = 0 (6.31)
σ_x = \frac{M_yz} {I_y} – \frac{M_zy} {I_z} Flexure Formula (6.30)
(a) Locate the neutral axis. From Eq. 6.33,
\tan β = \left(\frac{I_z}{I_y}\right) \tan θ (1)
where, from Fig. 1,
\tan θ = \frac{M_y} {M_z} = \frac{M \sin θ} {M \cos θ} (2)
Therefore,
\tan β = \left(\frac{305 in^4} {15.7 in^4}\right) \tan θ (3)
When θ = 0°, β = 0°, and the z axis is the neutral axis, but when θ = 2° we get
\tan β = \left(\frac{305 in^4} {15.7 in^4}\right) (0.0349) = 0.678
So, for θ = 2°,
β = 34.2° (a) (4)
The orientation of the NA at this cross section is shown in Fig. 2. (b) Determine the maximum tensile stress and maximum compressive stress. The maximum stresses will occur at the extreme points A and B. From Eq. 6.30,
σ_x = \frac{M_yz} {I_y} – \frac{M_zy}{I_z} (5)
σ_{xA} ≡ σ_x(6.0, -2.74) = \frac{(150 kip · in.)(\sin 2°)(-2.74 in.)}{15.7 in^4}-\frac{(150 kip · in.)(\sin 2°)(6.0 in.)}{305 in^4}
= -3.86 ksi
From symmetry,
σ_{xB} ≡ σ_x(-6.0, 2.74) = 3.86 ksi
Had there been no misalignment, that is, for θ = 0
σ_x(y, z) = \frac{-M_zy}{I_z} (6)
so, the maximum compression is at y = 6.0 in., or
(σ_x)_{max C} = \frac{-(150 kip · in.)(6.0 in.)} {305 in^4} = -2.95 ksi
and
(σ_x)_{max T} = 2.95 ksi
In summary,
Thus, a small misalignment of 2° increases the maximum stresses by 31%.
This is due to the large ratio of I_z/I_y.
Review the Solution The answers look reasonable, since a nonzero value of M_y, together with a large ratio of I_z/I_y, will cause a significant rotation of the neutral axis, that is, a large value of β. The values of the maximum stresses are not unreasonable.
(See Homework Problem 6.6-5 where the same loading is applied to a very similar sized beam, a W12×50.)
| θ | β | (σ_x)_{max C} | (σ_x)_{max T} |
| 0° | 0° | -2.95° ksi | 2.95 ksi |
| 2° | 34.2° | -3.86° ksi | 3.86 ksi |
