Determining Slope and Deflection in Beams

Calculate the slope and deflection at key points on various beams by analyzing moment of inertia, loadings, and material properties. Solutions include detailed calculations and diagrams to aid in understanding the structural behavior.

• Beams
• Slope
• Deflection
• Analysis
• Mechanics

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Uploaded on Feb 23, 2025 | 1 Views

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1. Example 8-7 Determine the slope and deflection at points B and C of the beam shown in the figure. Values for the moment of inertia of each segment are indicated in the figure. Take E = 200 GPa. A B C 500 N m IAB = 8(106) mm4 IBC = 4(106)mm4 4 m 3 m 1

2. SOLUTION 500 N m A B C 500 N m IAB = 8(106) mm4 IBC = 4(106)mm4 4 m 3 m IAB =2IBC = = the area under the M/EI diagram between A and B B B/A M /EIBC 500 250 EIBC 250 N m EI =( )(4m) BC EIBC x (m) (250 )(4) =1000 EIBC 100 EIBC N m2 = EIBC tangentB C B= B/A B/A A 100 N m2 B = tangentA (200 109 N )(4 10 6 m4 ) m2 B = 0.00125 rad 2

3. A B C 500 N m IAB = 8(106) mm4 IBC = 4(106)mm4 4 m 3 m IAB =2IBC = = the area under the M/EIdiagram between A and C =100 N m+1500 N m C C/A M /EIBC 500 250 EI 2 2 BC EIBC EIBC EIBC x (m) (250 )(4) =1000 EIBC (500 )(3) =1500 EIBC 2500 N m2 EIBC EIBC = EIBC 2500 N m2 C tangentC = C/A = A B 9N (200 10 )(4 10 m ) 6 4 tangentCA C/A m2 C = 0.00313 rad 3

4. A B C 500 N m IAB = 8(106) mm4 IBC = 4(106)mm4 4 m 3 m IAB =2IBC = t = the moment of the M/EI diagram between A and B computed about point B. B B/A M /EIBC 500 250 (250 )( 4) =1000 EIBC EI EIBC BC EI BC =1000 N m2 EIBC (2m) x (m) 2m =2000 N m3 EIBC tangentB C A B 2000 N m3 = tangent A (200 109N)(4 10 6m4) =tB/A B m2 B = 0.0025 m = 2.5mm 57

5. A B C 500 N m IAB = 8(106) mm4 IBC = 4(106)mm4 4 m 3 m IAB =2IBC = t = the moment of the M/EI diagram between A and C computed about point C. C C/A M /EIBC 500 500 1500 EI 250 )(3) = ( EIBC E I BC BC EI BC =1000 N m2 1500 N m2 (5 m)+ (1.5m) x (m) 1.5m EIBC EIBC (250 )(4) =1000 EIBC EIBC 7250 EIBC 5 m N m3 = tangentC C 7250 N m3 A = B N m2 tangentA (200 109 )(4 10 6m4) =t C/A C C = 0.00906 m = 9.06mm 5

6. Example 8-8 Determine the slope and deflection at points C of the beam shown in the figure. Take E = 200 GPa and I = 60(106) mm4 20 kN A B C D 3 m 3 m 6 m 6

7. SOLUTION ElasticCurve 20 kN 3 m 3 m 6 m tangentB A B A B C ' C D D tC/AC tangent C 10 10 kN 6 m 3 m 3 m tB/A M /EI 10(6)=60 /EI ' =tB /A 3 tangentA 12 x (m) ' =3t =1t 4 B/ A B/ A 12 A B C = D/CD C = ' tC /A horizontal C 1 4 tangentD = D/CtangentC t t (2) B/ A C /A C C = D/C --------(1) 7

8. Moment-AreaTheorem M /EI 6 m 6 m 1(30+60)(3) =135 60 /EI 2 EI EI EI 30 /EI M /EI x (m) A 60 /EI 1(6)(60) =180 2 EI 1(6)(60) =180 B 3 m C3 mD EI 6 m 2 EI EI x (m) A D/C = the area under the M/EI diagram between C and D C D B 26 =4 3 6+16 =8 135 kN m2 3 = D/C EI tB/A = the moment of the M/EI diagram between A and B computed about point B. M /EI 60 /EI 30 /EI x (m) A 180 kN m2 180 kN m2 tB / A =( )](4 m)+( )](8m) B D C EI EI 1 m =2160 kN m3 t = the moment of the M/EI diagram t C/A between A and C computed about pointC. B/ A EI =[1(3m)(30 kN m)](1m) =45kN m 2 EI t 3 C /A EI 8

9. Substitute D/C in(1): Substitute tB/A and tC/A in (2) : 1 C =4 tB / A tC / A =135 kN m2 = D/C C EI =1 ( 2160) 45 4 EI 135kN m 2 EI = (200 106kN)(60 10 6m4) 495 EI m2 = 495 = 0.01125rad = C (200)(60) C = 0.04125 m = 41.25mm, C = 41.25 mm A B C tangent C C = 0.01125 rad 9

10. Example 8-9 Determine the slope and deflection at points C of the beam shown in the figure. Take E = 200 GPa and I = 6(106) mm4 5 kN m A B C 4 m 4 m 10

11. ElasticCurve SOLUTION 5 kN m tangentB A B A B C C ' tangentC C tC/A 4 m 4 m tB/A tangentA M /EI 4 m 4 m 5 /EI x (m) tangentB ' =tB /A 4 8 A A B C C C/A 1 2 '= tangentC tB/A tangent A tB/ A C = ' tC /A =tB /A A 8 =1t C = A - C/A t (2) So; B/ A C /A C 2 =tB /A So; (1) C /A C 8 11

12. Moment-AreaTheorem M /EI M /EI 1(4)(5 2.5) =5 2 EI 2.5 /EI 1 2 5 20 EI ) = ( )(8)( 5 /EI 5 /EI EI EI EI B x(m) x (m) A B C 2 2 24 3 4 m A 2.5 EI(4)= 10 EI 8 3 8 m C/A = the area under the M/EI diagram between A and C tB/A = the moment of the M/EI diagram between A and B computed about point B. C /A=10+5=15 kN m2 EI EI EI (2 8m) 3 =20 kN m 2 tB/ A EI = the moment of the M/EI diagram t C/A between A and C computed about pointC. 106.667 kN m3 = 10 kN m2 5 kN m2 2 ( 4m) 3 EI (2 m)+ tB / A= EI EI =33.333 kN m3 EI 12

13. Substitute tB/A and C/A in (1): Substitute tB/A and tC/A in (2) : =tB /A 1 C =2 tB / A tC / A C /A C 8 =1 106.667 kN m3 ( 2 =1 106.667) (15) 33.333kN m3 ( ) C 8 EI EI EI EI = 1.667 kN m2 20 kN m3 = EI EI 20 kN m3 1.667 kN m2 (200 106kN)(6 10 6m4) = = (200 106kN)(6 10 6m4) m2 m2 C = 0.01667 m = 16.67 mm, = -0.0014rad C = 16.67 mm A B C= 0.0014 rad tangentC C 13

14. Example 8-10 Determine the slope and deflection at points C of the beam shown in the figure. Take E = 200 GPa and I = 250(106) mm4 1 kN/m B A C 8 m 8 m 14

15. ElasticCurve SOLUTION tangentA 1 kN/m tangentB B A tB/A C tC/A A C 4 kN 12 kN B 8 m 8 m C 8 m 8 m tangentC M /EI A B C x (m) '=tB /A 16 -32/EI 8 tangentA tangentC '= 2tB/ A tA/C A ' C = tC / A B C C C 2tB/ A (2) C = tC / A =tA /C+ C (1) C 16 15

16. Moment-AreaTheorem 32 32 1 3(8)( 85.333 EI 1 3(8)( 85.333 EI ) = ) = EI EI M /EI A M /EI A B B C C 8 m 8 m 8 m 8 m x (m) x (m) 1(8)( 32) = 128 2 EI 1(8)( 32) = 128 2 EI 3 4 EI EI 8 -32/EI -32/EI 28 3 8 3 8 1 4 8+ 8 tC/A = the moment of the M/EI diagram between A and C computed about point C. = ( 85.333)(38)+( 128)(8+18) EI 4 tA/C = the moment of the M/EI diagram between A and C computed about point A. )( 8)+( 85.33)(8+18) EI 3 = ( 128 2 t tA/C C /A 3 4 EI EI = 1877.37 kN m3 EI = ( 128)(18) EI = 1536 kN m3 EI Note b t 1 3 A= bh B/ A 3 h x =3b = 341.33kN m 3 x 4 EI 16

17. Substitute tC/A and B/A in (2): Substitute tA/C and C in (1) : 2tB/ A C = tC / A 1 = + ) (t A/C C C 16m =1877.37 2(341.33) EI 1 1536 kN m3 +1194.71kN m3 = ( ) EI C 16 m EI EI =1194.71kN m3 170.67 kN m2 = EI EI 170.67 kN m2 (200 106kN)(250 10 6m4) 1194.71kN m3 (200 106kN)(250 10 6m4) = = m2 m2 C = 0.02389 m = 23.89mm, C = 0.00341 rad A C= -23.89 mm = 0.00341rad C B C tangent C 17