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Chrome File Edit View History Bookmarks People Window Help a S A R 15% D Tue Aug 29 8:37:36 PM Q • O• R 2) Pinterest x 2 Statesboro, GA 1c x W Auto Logged out x 3 MasteringEngine ex 3 MasteringEnginee x C Civil Engineering ex D Mathway | Math P x O Caitlin ca Secure https:// session.masteringengineering.com /mycitemViewPassignmentProblemID=6257590&offset;=prev R & D O O Stress and Strain Problem 1.27 Resources D previous 9 of 10 | next > Problem 1.27 Part A Determine the resultant normal force acting on the cross section at B. The pipe has a mass of 10 kg/m and is fixed to the wall at A. Three forces, F . = 300 N, F = 420 N, and F = 515 N, act on the structure as shown. (Figure 1) Express your answer to three significant figures and include appropriate units. H WA sh o CD ? NB = Value Units Submit My Answers Give Up Figure 1 of 1 Part B Determine the resultant shear force acting in the T direction on the cross soction at B. Express your answer to three significant figures and include appropriate units. 2m CH sh a CD ? (VB). = Value | _ Units submit my answer she Up 2 m Part C 9 2 o® 0 E 0 OAA *with 29 oaS ERT

Explanation / Answer

Part A:

Normal force at B =F2 + (4/5)F3

= 420+(4/5)*515 = 832 N

Part B : Shear Force acting in the x direction at B = F1 = 300N

Part C : Shear force at B in z direction = (3/5)F3 + weight of pipe of portion BC = (3/5)*515 + (10*4*9.8) = 701 N

Part D : Resultant torque at B=F1*2 = 300*2=600Nm

Part E : Bending moment in x direction at B = F1*2 = 300*2=600Nm

Part F : Bending moment in z direction at B =(F2*2)+(3/5)*F3*2 + (10*2*9.8*2)+(10*2*9.8*1)

= (420*2)+(3/5)*(515*2)+(392)+(196)

= 1626Nm

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