Question

d = 25mm $d_3 = 42 \text{ mm}$ d? = 36 mm P? = 30 kN P? = TBD ? 1A B C brass rod rigid plate Steel pipe A steel pipe is attached to a brass rod by a rigid plate, as shown, and axial loads P? and P? are exerted at B and C respectively. If the allowable shear stresses in the brass and steel are $\tau_{allow, b} = 120 \text{ MPa}$ $\tau_{allow, s} = 200 \text{ MPa}$ find the max permissible value of P?.

          d = 25mm
$d_3 = 42 \text{ mm}$
d? = 36 mm
P? = 30 kN
P? = TBD
?
1A
B
C
brass rod
rigid plate
Steel pipe
A steel pipe is attached to a brass rod by a rigid plate, as shown,
and axial loads P? and P? are exerted at B and C respectively.
If the allowable shear stresses in the brass and steel are
$\tau_{allow, b} = 120 \text{ MPa}$
$\tau_{allow, s} = 200 \text{ MPa}$
find the max permissible value of P?.

d = 25mm
d3 = 42 mm
d? = 36 mm
P? = 30 kN
P? = TBD
?
1A
B
C
brass rod
rigid plate
Steel pipe
A steel pipe is attached to a brass rod by a rigid plate, as shown,
and axial loads P? and P? are exerted at B and C respectively.
If the allowable shear stresses in the brass and steel are
τallow, b = 120 MPa
τallow, s = 200 MPa
find the max permissible value of P?.

Added by Sharon R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Timothy James

Step 1

First, let's calculate the area of the steel pipe. The diameter of the pipe is given as 42 mm, so the radius would be half of that, which is 21 mm or 0.021 m. The area of a circle is given by the formula A = πr^2, so the area of the steel pipe is A = π(0.021)^2 = Show more…

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Thanks for your feedback!

42 mm x 9E=2p d = 25 mm = 30 kN rasrc at A steel pipe is attached to a brass rod by a rigid plate, as shown. The loads P and Pz are exerted at B and S respectively. 120 MPa o w aZ=