Mechanics_of_Materials_Chap_06-02 - SECTION 6.4 Beams with...

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Beams with Inclined Loads When solving the problems for Section 6.4, be sure to draw a sketch of the cross section showing the orientation of the neutral axis and the locations of the points where the stresses are being found. Problem 6.4-1 A beam of rectangular cross section supports an inclined load P having its line of action along a diagonal of the cross section (see figure). Show that the neutral axis lies along the other diagonal. Solution 6.4-1 Location of neutral axis SECTION 6.4 Beams with Inclined Loads 391 b z y C P h Load P acts along a diagonal. I z I y 5 h 2 b 2 I y 5 hb 2 12 I z 5 bh 3 12 tan a 5 b / 2 h / 2 5 b h See Figure 6-15b. b 5 angle between the z axis and the neutral axis nn u 5 angle between the y axis and the load P u 5 a 1 180º tan u 5 tan ( a 1 180º) 5 tan a (Eq. 6-23): [ The neutral axis lies along the other diagonal. QED 5 ¢ h 2 b 2 ¢ b h 5 h b tan b 5 I z I y tan u 5 h 2 b 2 tan u b z y C P h n n a b Problem 6.4-2 A wood beam of rectangular cross section (see figure) is simply supported on a span of length L . The longitudinal axis of the beam is horizontal, and the cross section is tilted at an angle a . The load on the beam is a vertical uniform load of intensity q acting through the centroid C . Determine the orientation of the neutral axis and calculate the maxi- mum tensile stress s max if b 5 75 mm, h 5 150 mm, L 5 1.5 m, a 5 30°, and q 5 6.4 kN/m. C z y q h b a Probs. 6.4-2 and 6.4-3
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Solution 6.4-2 Simple beam with inclined load 392 CHAPTER 6 Stresses in Beams L 5 1.5 m q 5 6.4 kN/m b 5 75 mm h 5 150 mm a 5 30º B ENDING MOMENTS 5 900 N ? m 5 1559 N ? m M z 5 q y L 2 8 5 q (cos a ) L 2 8 M y 5 q z L 2 8 5 q (sin a ) L 2 8 M OMENTS OF INERTIA N EUTRAL AXIS nn (E Q . 6-23) b 5 66.6º M AXIMUM TENSILE STRESS ( AT POINT A ) (E Q . 6-18) s max 5 11.9 MPa 1 (1559 N ? m)(75 mm) 21,094 3 10 3 mm 4 5 (900 N ? m)(37.5 mm) 5273 3 10 3 mm 4 s max 5 M y ( b / 2) I y 2 M z ( 2 h / I z 5 ¢ h b 2 tan a 5 4 tan 30 85 2.3094 tan b 5 I z I y tan u 5 I z I y tan a I z 5 bh 3 12 5 21,094 3 10 3 mm 4 I y 5 hb 3 12 5 5,273 3 10 3 mm 4 C z y q h b a n n b A Problem 6.4-3 Solve the preceding problem for the following data: b 5 6 in., h 5 8 in., L 5 8.0 ft, tan a 5 1/3, and q 5 375 lb/ft. Solution 6.4-3 Simple beam with inclined load L 5 8.0 ft q 5 375 lb / ft b 5 6 in. h 5 8 in. tan a 5 1 / 3 B ENDING MOMENTS 5 11,380 lb-in. M y 5 q z L 2 8 5 q (sin a ) L 2 8 cos a 5 3 Ï 10 sin a 5 1 Ï 10 q a n n b A 5 34,150 lb-in. M OMENTS OF INERTIA N EUTRAL AXIS nn (E Q . 6-23) b 5 30.7º M AXIMUM TENSILE STRESS ( AT POINT A ) (E Q . 6-18) s max 5 M y ( b / I y 2 M z ( 2 h / I z 5 771 psi 5 ¢ h b 2 tan a 5 0.5926 tan b 5 I z I y tan u 5 I z I y tan a I z 5 bh 3 12 5 256 in. 4 I y 5 hb 3 12 5 144 in. 4 M z 5 q y L 2 8 5 q (cos a ) L 2 8
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Problem 6.4-4 A simply supported wide-flange beam of span length L carries a vertical concentrated load P acting through the centroid C at the midpoint of the span (see figure). The beam is attached to supports inclined at an angle a to the horizontal.
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Mechanics_of_Materials_Chap_06-02 - SECTION 6.4 Beams with...

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