Question

Aluminum Copper Steel $E_{al} = 10(10^3) \text{ ksi}$ $E_{cu} = 18(10^3) \text{ ksi}$ $E_{st} = 29(10^3) \text{ ksi}$ 3.50 kip 1.75 kip 2.00 kip 1.50 kip A B 3.50 kip C 1.75 kip D -18 in. -12 in. -16 in-

          Aluminum
Copper
Steel
$E_{al} = 10(10^3) \text{ ksi}$  $E_{cu} = 18(10^3) \text{ ksi}$  $E_{st} = 29(10^3) \text{ ksi}$ 
3.50 kip
1.75 kip
2.00 kip
1.50 kip
A
B 3.50 kip C 1.75 kip D
-18 in. -12 in. -16 in-

Aluminum
Copper
Steel
Eal = 10(10^3) ksi Ecu = 18(10^3) ksi Est = 29(10^3) ksi
3.50 kip
1.75 kip
2.00 kip
1.50 kip
A
B 3.50 kip C 1.75 kip D
-18 in. -12 in. -16 in-

Added by Lee C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suman Saurav Thakur

Step 1

Part A: To determine the average normal stress in the aluminum section, we need to calculate the total force acting on the aluminum section and divide it by the cross-sectional area of the aluminum section. Show more…

Show all steps

Thanks for your feedback!

Problem 4.3 The composite shaft, consisting of aluminum, copper, and steel sections, is subjected to the loading shown. The cross-sectional areas of sections AB, BC, and CD are A_AB = 0.08 in^2, A_BC = 0.15 in^2, and A_CD = 0.07 in^2, respectively. The modulus of elasticity for each section is shown in the figure. Neglect the size of the collars at B and C. Part A Determine the average normal stress in the aluminum due to this loading. Part B Determine the average normal stress in the steel due to this loading. Aluminum: E = 10(10^3) ksi Copper: E_cu = 18(10^3) ksi Steel: E_s = 29(10^3) ksi Loadings: A = 3.50 kip B = 1.75 kip C = 2.00 kip D = 1.50 kip Lengths: AB = 13.50 in BC = 12 in CD = 16 in D = 18 in