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*12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6 (10³)ksi.

*12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6 (10³)ksi.

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter9: Deflections Of Beams
Section: Chapter Questions
Problem 9.9.4P: The cantilever beam shown in the figure supports a triangularly distributed load of maximum...
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Figure 12-47 Full Alternative Text *12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew 1.6 (10³)ksi. Prob. 12-48 A -9 ft 0.8 kip/ft B = 9 ft 1.5 kip C HE 6 in. 12 in.
Transcribed Image Text:Figure 12-47 Full Alternative Text *12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew 1.6 (10³)ksi. Prob. 12-48 A -9 ft 0.8 kip/ft B = 9 ft 1.5 kip C HE 6 in. 12 in.
3.6 6 3 9 Ay By EMA=0= -(6.3.6) + (9. By) -(18.1.5) By = (6.3.6) + (18.1.5) = 5.4 Kip ΣFy=0= + Ay -3.6 +5.4 -1.5 Ay= 0.3 Kip R = = (x₁) (0-0898, xJ (0.0888 x₁) XV ^^ 0.3 X₁ EMO=O=M + M = -0.08x₁³ 0.3 X₁ El dx = -0.0888x²³ -0.3X XI 0-0888X, 2 -0.0888 x 24 -0.0888X,5 120 0.8 0.8 = # H X₁ 2 0.3x₁² (X₁.0.3) Kip-ft + Ci El V=SS -0.0888 X1² - 0.3 X₁² + C₁ 24 2 3 0.3 X+0 6 X₁ (1 + CIXI + C₂ H M 1.5 (18-X2) EMO =O=-M-((18-x₂)-1.5) =-M-(27-1.5 X₂) M= 1.5 X2-27 El=1.5 X₂-27 ↳1.5 X2-27X2 + C3. 6 Ev= 1.5x23 6 3 El ETV = S 15X²²-27X2² - C3X= + C4 4 1.5x₁ + 2 Elv = -0.0988X5 120 El = 15x₂² - 27 x₂ -C36 X1=0 El ay = -0.0888x₁¹_ 0.3x² + ₁ 我 24 V₁=0 ²-27 x2² - (3X² + C+ +C4 (4) - 0.3 x₁ ³ + Cixi + C₂ ☺ (2) C₂=0 @ 0=-0.0888 (915-03 (95²- C₁ (1)-0 + X₁=9| X₁=9 Vi=O 6 80.146 = C1 (a) - Ci=8.905 dvi-dvz dx dx
Transcribed Image Text:3.6 6 3 9 Ay By EMA=0= -(6.3.6) + (9. By) -(18.1.5) By = (6.3.6) + (18.1.5) = 5.4 Kip ΣFy=0= + Ay -3.6 +5.4 -1.5 Ay= 0.3 Kip R = = (x₁) (0-0898, xJ (0.0888 x₁) XV ^^ 0.3 X₁ EMO=O=M + M = -0.08x₁³ 0.3 X₁ El dx = -0.0888x²³ -0.3X XI 0-0888X, 2 -0.0888 x 24 -0.0888X,5 120 0.8 0.8 = # H X₁ 2 0.3x₁² (X₁.0.3) Kip-ft + Ci El V=SS -0.0888 X1² - 0.3 X₁² + C₁ 24 2 3 0.3 X+0 6 X₁ (1 + CIXI + C₂ H M 1.5 (18-X2) EMO =O=-M-((18-x₂)-1.5) =-M-(27-1.5 X₂) M= 1.5 X2-27 El=1.5 X₂-27 ↳1.5 X2-27X2 + C3. 6 Ev= 1.5x23 6 3 El ETV = S 15X²²-27X2² - C3X= + C4 4 1.5x₁ + 2 Elv = -0.0988X5 120 El = 15x₂² - 27 x₂ -C36 X1=0 El ay = -0.0888x₁¹_ 0.3x² + ₁ 我 24 V₁=0 ²-27 x2² - (3X² + C+ +C4 (4) - 0.3 x₁ ³ + Cixi + C₂ ☺ (2) C₂=0 @ 0=-0.0888 (915-03 (95²- C₁ (1)-0 + X₁=9| X₁=9 Vi=O 6 80.146 = C1 (a) - Ci=8.905 dvi-dvz dx dx
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