# 36 3 3333 5 1 17920 it vq psi 450 1 3333 5 5 1 2 1 18

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36 . 3 3333 . 5 1 17920 It VQ psi 450 1 3333 . 5 5 . 1 1600 10 1600 2 1 18 2880 2 1 M M=17.92 k.in A y in 1 in 4 in 1

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8 Example 48: Consider the cantilever beam subjected to the concentrated load shown below. The cross section of the beam is of T-shape. Determine the maximum shearing stress in the beam and also determine the shearing stress 25 mm from the top surface of the beam of a section adjacent to the supporting wall. 0 A M M-50×2=0 M=100 KN.m 0 y F A y -50=0 A y =50 KN S.F. Diagram B.M. Diagram From shear and bending moment diagrams V=50 KN M=100 KN.m A B KN 50 m 2 mm 200 mm 50 mm 50 mm 125 M B KN 50 m 2 y A m KN . 100 B KN 50 m 2 KN 50 KN 50 m KN . 100 mm 200 mm 50 mm 50 mm 125 1 2
9 A A y y c mm m y c 65 . 58 10 65 . 58 10 16250 10 953125 3 3 9 I=I 1 +I 2 2 3 1 12 Ad bh I 2 3 6 3 3 3 1 ) 10 65 . 33 ( 10 10000 12 ) 10 50 ( 10 200 I I 1 =13.40655833×10 -6 m 4 2 3 2 12 Ad bh I 2 3 6 3 3 3 2 ) 10 85 . 53 ( 10 6250 12 ) 10 125 ( 10 50 I I 2 =26.26191146×10 -6 m 4 I=39.6684×10 -6 m 4 Q A y A =50×10 -3 ×116.35×10 -3 =0.0058175 m 2 y =58.175×10 -3 m Q=0.000338433 m 3 It VQ max MPa 5315553 . 8 10 50 10 6684 . 39 000338433 . 0 10 50 3 6 3 max No. of Area A(m 2 ) y (m) A y (m 3 ) 1 10000×10 -6 25×10 -3 250000×10 -9 2 6250×10 -6 112.5×10 -3 703125×10 -9 A =16250×10 -6 A y =953125×10 -9 mm 200 mm 50 mm 50 mm 125 1 2 A N . mm c 65 . 58 2 mm c 35 . 116 1 mm 200 mm 50 mm 50 mm 125 1 2 A N . mm c 65 . 58 2 mm c 35 . 116 1 A y

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0 Q A y A =50×10 -3 ×25×10 -3 =0.00125 m 2 y =103.85×10 -3 m Q=0.000129812 m 3 It VQ MPa 272441 . 3 10 50 10 6684 . 39 000129812 . 0 10 50 3 6 3 mm 200 mm 50 mm 50 mm 125 1 A N . mm c 65 . 58 2 mm c 35 . 116 1 A y mm 25
1 Curved Beams Due to the curvature of the beam, the normal strain in the beam does not vary linearly with depth as in the case of a straight beam .As result, the neutral axis does not pass through the centroid of the cross section. C7 If we isolate a differential segment of the beam let a strip material located at r distance has an original length r dθ .Due to the rotations δθ/2, the strip's total change in length is δθ)R-r( Strain is a nonlinear function of r, in fact it varies in a hyperbolic fashion .Hooke's law applies, 0 R F A A dA r r R Ek dA 0 ) ( 0 A A dA r dA R 0 ) ( , ) ( r r R k d k rd r R ) ( r r R Ek

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2 ) ( y R Ae My R : - The location of the neutral axis, specified from the center of curvature 0' of the member. A : - The cross -sectional area of the member. r : - The arbitrary position of the area element dA on the cross section, specified from the center of curvature 0' of the member. y=R -r , e= r -R σ : - The normal stress in the member. M : - T he internal moment, determined from the method of sections equations of equilibrium and computed about the centroidal axis .
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• Winter '15
• MAhmoudali

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