Question

Part II: Show the detail steps of the following calculation Problems 2.5. 2. (30 pts) A 3-in cast iron plate (E - 10.4 Mpsi) and a 2 in Copper plate - 17.2 Mpi) we compressed with one bolt and one regular hexagonal mut. The bolt is 1/4 in 20 UNC. (2.1) (4 pts) Determine a suitable length for the bolt, rounded up to the nearest ain (2.2) (8 pts) Determine the carbon steel (E - 30.0 Mpsi) bolt's stiffness (2.3) (18 pts) Determine the stiffness of the members, ko 3. (17 nte) Savannlaan cin. . . .. .

Answer 1

PART T Soln Given Data : thickness of cast iron Plate, tc = 3 in thickness of copper Plate, t copper - 2 in The bolt is 1/4 in -20 UNG. (Nominal dia 2 1/4 in) Young's modulus for cast iron Plate, E= 10.4 mesi Young's modulus foc compet Plate, E- 17.2 mesi (2.1) For nominal size of 1/4 in find height H for hexagonal nut from table A-31 (Shigley Design gthed.) He 7/99 in -0.21875 in Compute the grip length & l = tc t t Copper b- (3 in) + (2 in) e = 5 in compute the length of the bolt: La lt H La (5 in) +0.21875 in ) L = 5.21875 in 5.3 in Ans . Therefore, the suitable length of the bolt 5.30

find out by using - (2.2) The bolt's stiffness can be following expression: khe . Ad At Adlt + E At ld where, Kb = bolt's stiffness. Id = length of unthreaded portion in geiß lt = length of threaded portion in grip Ad - Area of unthreaded portion At - Area of threaded portion E = Young's modulus - 30 MPsi Threaded length (LT); For Le 6 in - LT = (20 + 1/4) in LT = 2(0.25 + 1/4 Lt = 0.75 in ld = L-LT - (5.3 in ) - 10.75 in) - 4.55 in lt = l-ld = (5 in ) - (4.55 in) = 0.45 in Ad = T d ²/4 = T (0.25)² - 0.049 in2 e 20 UNC At = 0.0318 in² from table 8-2) 1 & 1² in so, kb (0.049) Co.0318) (30) (0.049) (0.45) +0.0318)(4:55) K6 = 0.28 M16b/in/ Ane.

the First find the individual stiffness for Cast iron and copper Plate by using ; . in 0.577 4 Tu Stod 55+ to-d) CD + d ) (1.255€ totd) (D-d . Then , using individual stiffness kig ko ... km you can find out the member stiffness. - km - 1 tltt tat Iki K2 K3 Ki