# Ft 22 k p 12 ft k q 5 1 b of the cross sectional s a

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ft 22 k P 12 ft k q / 5 . 1 B of the cross sectional s a uniform load of rm load includes an a point 9 ft from the ated wood and has a maximum tensile and

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6 0 A M B y ×22-12×9-33×11=0 B y =21.409 k 0 y F A y +21.409-12-33=0 A y =23.591 k S.F. Diagram B.M. Diagram k 12 k 33 y A y B ft 9 k P 12 ft k q / 5 . 1 k 591 . 23 k 409 . 21 k 591 . 23 k 409 . 21 k 091 . 10 k 909 . 1 ft k . 569 . 151 Maximum bending moment M max =151.569 k.ft =151.569×12 =1818.828 ksi
7 Example 44: The simply supported beam has the cross sectional area shown below. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location. in 75 . 8 in 27 1 c 2 c c 1 =c 2 =13.5 in 1 = I Mc 1 I= 12 3 bh = 4 3 1875 . 14352 12 ) 27 ( 75 . 8 in 1 = 1875 . 14352 5 . 13 10 828 . 1818 3 =-1710.8317 psi 1 = 1875 . 14352 5 . 13 10 828 . 1818 3 =1710.8317 psi A B m KN / 5 m 6 mm 250 mm 300 mm 20 mm 20 mm 20 KN 30 y A y B

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8 0 A M B y ×6-30×3=0 B y =15 KN 0 y F A y +15-30=0 A y =15 KN N.A c 1 =c 2 =170 mm I 1 = 2 3 12 Ad bh I 1 = 2 3 3 3 3 3 3 ) 10 160 ( ) 10 20 10 250 ( 12 ) 10 20 ( 10 250 I 1 =128.16667×10 -6 m 4 I 3 =I 1 =128.16667×10 -6 m 4 I 2 = 12 3 bh = 12 ) 10 300 ( ) 10 20 ( 3 3 3 =45×10 -6 m 4 I= I 1 + I 2 + I 3 =128.16667×10 -6 +128.16667×10 -6 +45×10 -6 I=301.333×10 -6 m 4 max = I Mc 1 = 6 3 3 10 333 . 301 10 170 10 5 . 22 =12.693598 MPa. m KN / 5 KN 15 KN 15 KN 15 KN 15 m KN . 5 . 22 Maximum bending moment M max =22.5 KN.m mm 250 mm 300 mm 20 mm 20 mm 20 1 2 3
9 B = I My B = 6 3 3 10 333 . 301 10 150 10 5 . 22 =11.200233 MPa. Example 45: The beam shown below has a cross section of channel shape with width b=300 mm and height h=80 mm , the web thickness is t=12 mm . Determine the maximum tensile and compressive stresses in the beam due to uniform load. 0 A M B y ×3-14.4×2.25=0 B y =10.8 KN 0 y F A y +10.8-14.4=0 A y =3.6 KN 0 x F A x =0 B m 3 m 5 . 1 m KN / 2 . 3 A B mm 300 mm 80 mm 12 x A y A KN 4 . 14 y B m KN / 2 . 3

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0 M 1 =2.025 KN.m M 2 =3.6 KN.m A A y y c 6 9 10 5232 10 321888 c y =61.52×10 -3 m y c =61.52 mm I 1 = 2 3 12 Ad bh No. of Area A(m 2 ) y (m) A y (m 3 ) 1 960×10 -6 40×10 -3 38400×10 -9 2 3312×10 -6 74×10 -3 245088×10 -9 3 960×10 -6 40×10 -3 38400×10 -9 A =5232×10 -6 A y =321888×10 -9 KN 6 . 3 KN 8 . 10 KN 6 . 3 KN 6 KN 8 . 4 m KN . 025 . 2 m KN . 6 . 3 mm 300 mm 80 mm 12 1 2 3 mm 80 mm 12 mm 300 mm c 52 . 61 1 mm c 48 . 18 2
I 1 = 12 ) 10 80 ( 10 12 3 3 3 + I 3 = I 1 =0.95658×10 -6 m I 2 = 2 3 12 Ad bh I 2 = 12 ) 10 12 ( 10 276 3 3 3 I= I 1 + I 2 + I 3 =2.46874× 46 . 2 10 025 . 2 ) ( 2 1 1 I c M t 468 . 2 10 6 . 3 ) ( 3 1 2 2 I c M t ( t ) max =50.462179 MP 2 025 . 2 ) ( 1 1 1 I c M c 2 1 6 . 3 ) ( 2 2 2 I c M c ( c ) max =-89.71054 MP Composite Beams: C omposite beams are m load. Normal stress in material 1 is de Normal stress in material 2 is de dA=dydz The force dF acting on the area dF= dA=( E 1 ε)dydz If the material 1 is being transfo b 2 =nb 1 +960×10 -6 ×(21.52×10 -3 ) 2 =0.95658×10 - 4 +3312×10 -6 ×(12.48×10 -3 ) 2 =0.55558×1 ×10 -6 m 4 462179 . 50 10 6874 10 52 . 61 0 6 3 3 MPa 94815 . 26 10 874 10 48 . 18 6 3 MPa Pa 158339 . 15 10 46874 .

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• Winter '15
• MAhmoudali

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