{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

02-03ChapGere

# 02-03ChapGere - 106 CHAPTER 2 Axially Loaded Members...

This preview shows pages 1–4. Sign up to view the full content.

106 CHAPTER 2 Axially Loaded Members Problem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1 / 8 in. Before they were loaded, all three wires had the same length. What temperature increase T in all three wires will result in the entire load being carried by the steel wires? (Assume E s 30 10 6 psi, s 6.5 10 6 /°F, and a 12 10 6 /°F.) Solution 2.5-3 Bar supported by three wires W = 750 lb S A S S steel A aluminum W 750 lb E s 30 10 6 psi E s A s 368,155 lb s 6.5 10 6 / F a 12 10 6 / F L Initial length of wires 1 increase in length of a steel wire due to temperature increase T s ( T ) L A s d 2 4 0.012272 in. 2 d 1 8 in. 2 increase in length of a steel wire due to load W /2 3 increase in length of aluminum wire due to temperature increase T a ( T ) L For no load in the aluminum wire: 1 2 3 or Substitute numerical values: N OTE : If the temperature increase is larger than T , the aluminum wire would be in compression, which is not possible. Therefore, the steel wires continue to carry all of the load. If the temperature increase is less than T , the aluminum wire will be in tension and carry part of the load. 185 F ¢ T 750 lb (2)(368,155 lb)(5.5 10 6 F) ¢ T W 2 E s A s ( a s ) s ( ¢ T ) L WL 2 E s A s a ( ¢ T ) L WL 2 E s A s W S A S Rigid Bar S A S W 2 W 2 3 1 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
SECTION 2.5 Thermal Effects 107 Problem 2.5-4 A steel rod of diameter 15 mm is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure. Calculate the temperature drop T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use 12 10 6 /°C and E 200 GPa.) Solution 2.5-4 Steel rod with bolted connection 15 mm 12 mm diameter bolt R rod B bolt P tensile force in steel rod due to temperature drop T A R cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P EA R ( T ) Bolt is in double shear. V shear force acting over one cross section of the bolt average shear stress on cross section of the bolt A B cross-sectional area of bolt t V A B EA R ( ¢ T ) 2 A B V P 2 1 2 EA R ( ¢ T ) S UBSTITUTE NUMERICAL VALUES : 45 MPa d B 12 mm d R 15 mm 12 10 6 / C E 200 GPa ¢ T 24 C ¢ T 2(45 MPa)(12 mm) 2 (200 GPa)(12 10 6 C)(15 mm) 2 ¢ T 2 t d B 2 E d R 2 A R d R 2 4 where d R diameter of steel rod A B d B 2 4 where d B diameter of bolt Solve for ¢ T : ¢ T 2 t A B EA R 15 mm 12 mm diameter bolt B R Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase T at distance x from end A is given by the expression T T B x 3 / L 3 , where T B is the increase in temperature at end B of the bar (see figure). Derive a formula for the compressive stress c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion .) L A T T B B x 0
108 CHAPTER 2 Axially Loaded Members Solution 2.5-5 Bar with nonuniform temperature change At distance x : R EMOVE THE SUPPORT AT END B OF THE BAR : Consider an element dx at a distance x from end A .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

02-03ChapGere - 106 CHAPTER 2 Axially Loaded Members...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online