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Unformatted text preview: Problem 4.3. For sections in Figures 4.20 and 4.21 the symmetry means that we have bending only about the y‐axis (same direction as the moment), so that one can use Eqs. 4.31 and 4.32. For the section in Fig. 4.22, the absence of symmetry means that you need to use Eq. 4.33. a. You first want to calculate the moments of inertia (answers: Iy=4Ah2, Iz=2Ah2, Iyz=‐ 2Ah2). b. Then you want to calculate the position of the neutral axis by setting the bending stress to zero in Eq. 4.33. This will allow you to see what point in the cross‐section is furthest away from the neutral axis and therefore will have the highest stresses. c. Finally you can compare the maximum stress in each one of the three sections and conclude which section is most effective (has the lowest stresses). The stringer‐ web sections shown in figures are subject to shear V z 0 , while Vy 0 . Find the bending stress in the stringers for the same bending moment M y . Which section is most effective in bending? Solution: (a) 2A z c 2h
2A y Given, V z 0 M Y 0 I Y 0, I YZ 0 (symmetry) Axial stress due to Moment My xx MY h Iy For the section (a) , I y 2 Ah 2 2 Ah 2 4 Ah 2 . Therefore xx (b) A MY (1) 4 Ah A 2h
y A A Iy 0 I yz 0 (Section is symmetric with respect to y) I y Ah 2 Ah 2 Ah 2 Ah 2 4 Ah 2 . Then, xx (c) A MY h MY (2) Ih 4 Ah h A z
2h y A h Lets find the centroid of the section (origin of the yz system), yc y A A
i i i hA 0 0 hA 0 4A zc 0 So, y & z is at centroid and yz yz . I yz Ai y i z i A( h)(h) A(0)(h) A(0)( h) A(h)( h) 2 Ah 2 I y Ai z i2 4 Ah 2 I z Ai y i2 2 Ah 2 From equation 4.29 (with M z 0 ) xx I yz M Y I y I z I yz 2 y IzMY I y I z I yz 2 z z xx M y = 2 Ah 2 2 Ah 2 y 2 4 24 24 8 A h 4 A2 h 4 8 A h 4 A h M y y z (3) 2 Ah 2 2 Neutral Axis for the section (c) and moment My can be found as tan If we plot this on the section, I yz M y IzM y 1 45 . 1 h 2 45 4 3 h Magnitude of the stresses (absolute value of the stress) will be identical at points 2 and 3. Stresses at 1 and 4 are zero since the neutral axis passes through them. Therefore it is sufficient to check points 2 and 3 for the maximum magnitude of the stresses. Axial stress at points (1) & (4) xx 0 Axial stress at points (2) & (3) xx Comparing (1), (2) and (4) we can conclude that: ‐ Section (b) and (a) has same effectiveness for bending. ‐ Section (c) is the least effective as the xx is the largest although same amount of material is used in all (a), (b) and (c). Problem 4.5: Figure shows the cross‐section of a four‐stringer box‐beam. Assume that the thin walls are ineffective in bending and the applied bending moments are 1 My (4) 2 Ah M y 500 KN .cm M z 200 KN .cm Find bending stresses in all stringers. 4 cm2 1 50 cm z 1 cm2 c y 4 3 200 cm 2 cm2 z
2 4 cm2 50 cm y yc y A A
i i i (200)(2) (200)(1) 54.55cm (4 4 2 1) (100)(4) (50)(1) 40.91cm (11) zc z A A
i i i I y Ai z i 2 (4)(59.09) 2 (4)(40.91) 2 (2)(40.91) 2 (1)(9.09) 2 I y 24090.91cm 4 I z Ai yi 2 (4)(54.55) 2 (4)(54.55) 2 (2 1)(145.45) 2 I z 87272.73cm 4 I yz Ai yi z i (4)(54.55)(59.09) (4)(54.55)(40.91) (2)(145.45)(40.91) (1)(145.45)(9.09) I yz 14545.45cm 4 xx I y M z I yz M y I y I z I yz
2 y I z M y I yz M z I y I z I yz
2 z stringer (1) stringer (2) stringer (4) stringer (3) y 54.55 xx 1201.9 N / cm 2 z 59.09 y 54.55 xx 951.95 N / cm 2 z 40.91 y 145.45 xx 384.59 N / cm 2 z 9.09 y 145.45 xx 692.33N / cm 2 z 40.91 Check: To check for an error we sum the axial forces on the stringers and make sure that they are zero: ‐1201.914+951.954‐384.591+692.332=0.23 0 Problem 4.6: From the data in the previous problem we obtain xx 1.2981 y 21.5385z So the neutral axis, obtained by setting the stress to zero is z=‐0.0603y. ...
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This note was uploaded on 11/29/2010 for the course AE 350 taught by Professor Achuthan during the Fall '09 term at Clarkson University .
 Fall '09
 ACHUTHAN

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