MasteringEngineering MasteringComputerSciences HW 3 Normal Stress - Google Chrom

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MasteringEngineering MasteringComputerSciences HW 3 Normal Stress - Google Chrome Secure https://session.mastenngengineering.com/myotitemView?assignment HW3 Normal Stess Problem 1.55 Proble mlD- 6266457 c previous | 3of4 I nexd Problem 1.55 Part A The 2 0-Mg concrete pipe has a center ot mass at point G, and is suspended trom cables AB and AC. The diameters of AB and AC are 12 mm and 10 mm. (Figure 1) Determine the average normal stress developed in the cable AB Express your answer to three significant figures and include the case of tension. appropriate units. Enter negative value in the case of compr ession and positive value in the OAB Value Units Submit My Answers Give Up Part B Delermine the average normal stress developed in the cable AC of 1 Express your answer to three significant fqures and include the appropnate units. Enter negative value in the case of compression and positive value in thre case of tension. 30 A0 : Value Units My Answers Sive Lle Submit ProvideFeedba Show all Masteringngin...t acer

Explanation / Answer

Mass = 2 Mg = 2 x 10^6 g = 2 x 10^3 kg = 2000 kg

Let Tb be the tension in AB

Let Tc be the tension in AC

Sum horizontal forces to zero, Right is positive direction

Tcsin45 - Tbsin30 = 0

Tc = Tbsin30/sin45

Sum vertical forces to zero, Up is positive

Tbcos30 + Tccos45 - 2000(9.81) = 0

Tbcos30 + (Tbsin30/sin45)cos45 = 2000(9.81)

Tb(cos30 + sin30cot45) = 2000(9.81)

cot45 = 1

Tb = 2000(9.81) / (cos30 + sin30)

Tb = 14362.84 N

Tc = Tbsin30/sin45

Tc = 14362.84sin30/sin45

Tc = 10156.06 N

= F/A

Part A

for AB

= 14362.84 / (0.012²/4)

= 127 x 10^6 Pa or 127 MPa

Part B

for AC

= 10156.06 / (0.010²/4)

= 129 x 10^6 Pa or 129 MPa

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