Determine the average normal stress developed in the cable A

Determine the average normal stress developed in the cable AB. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension.

Solution

Let us say : It is a 1.5-Mg concrete pipe and the diameters of AB and AC are 12 mm and 10 mm

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 - 1500(9.81) = 0
Tbcos30 + (Tbsin30/sin45)cos45 = 1500(9.81)
Tb(cos30 + sin30cot45) = 1500(9.81)
cot45 = 1
Tb = 1500(9.81) / (cos30 + sin30)
Tb = 10772 N

Tc = Tbsin30/sin45
Tc = 10772sin30/sin45
Tc = 7617 N

= F/A

for AB

= 10772 / (0.012²/4)
= 95245391 N/m²
= 95 MPa

for AC

= 7617 / (0.010²/4)
= 96983222 N/m²
= 97 MPa