Using whatever method you choose, find the force in member BG. 20 kips D 40 kips 18' – 0" E 15 kips 3 bays @ 15' - 0" = 45' – 0"
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Q: Using whatever method you choose, find the force in member BG. 20 kips
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- (7) The drive of an outboard motor which rotates @ 5,000rpm is supported by a spline in the crankshaft @ the upper end & bearing @ the lower end 560mm apart. (a) If the engine develops 15kW @ this speed, what is the diameter of the solid steel shaft to use? The allowable torsional stress is 80N/mm2 .. A tensile load of 1120 lbs is applied to a square bar,1/2 mm on a side and having a length of 5.4 ft. Compute the stress and the axial deformation in the bar if it is made from (a) AISI C1045 hot-rolled steel, (b) SAE 8640 OQT 1000 F steel.A round cross-sectional bar d= 39 mm is cut from steel of Sut = 715 MPa. The bar is loaded with a completely reversed stresses. What is the strength of failure (in MPa) If only 10 4 cycles are needed? (Take fully corrected Se = 358 MPa, no need to use modification fcactors) (accepted answer range +/- 5 MPa)
- question: a)Calculate the axial elongation of the member b)Calculate the new cross section area after the elongation ,considering the Poisson ration effect c)If the material has an ultimate shear strength of 12Mpa and applying a factor of safety of 1.5,what is the maximum tensile load this member could take without failing in shear.Calculate the failure plan inclination thetaThe flange coupling for two shafts made of plain carbon steel to be connected. The shafts transmitting a toque of 600 N-m. Assume the following permissible stresses for the coupling components: Shaft — Permissible shear stress = 35 MPa Keys — Rectangular formed end sunk key having permissible compressive strength = 60 MPa Bolts — Six numbers made of steel having permissible shear stress = 28 MPa Pitch circle diameter of bolts = 1.6 × Diameter of shaft 1. diameter of shaft =44.36 2. Hub length =67.5 3. Hub diameter 90 mm Calculate: 2. thickness of flange; 3. Diameter of flange 4. Width of key 5. Thickness of key 5. Diameter of bolt; 6. Crashing stress in boltA round bar, 40 mm in diameter, has a shallow circumferential notch with a depth c = 2 mm with a root radius r = 50 microns. The bar is made of a low carbon steel with a yield strength of σy = 300 MPa. It is loaded axially with a nominal stress, σnom. At what value of σnom will yield first commence at the root of the notch?
- 18. A steel bar is desired to be formed until it has a final length of 1.8m. What length should the bar in m be before the forming stress is removed? The yield strength is 427 MPa, the tensile strength is 572 MPa, and the youngs modulus 210 GPa. Round of to tenths.Given: angle α=64o. Cross sections of both rods are square with sizes a1=21mm, a2=13mm. The rods are made of steel with yielding stress σY=270MPa; factor of safety n=2,2. Find: allowable value of force (in kN) from strength calculation FA = ?Given a 1.3meter Wide Flange steel compression member with A=4,118mm2, Ix=87,958,354mm4 and Iy=3,574,253mm4 with Fy = 345MPa. Calculate the critical buckling stress in MPa if both ends are fixed or rigid. Express your answer in 2 decimal places.
- Calculate each of the following under 20 kN load on a structure given below where all pins have a diameter of 10 mm. a) What is maximum stress on bar BD? b) Check the Pin strength at Location D against shearing. (Use ?Y=300 MPa, n=2 - factor of safety for Pin). Is it secure/safe or not? c) Check the Pin and Bearing strength against shearing at Location C. Are they secure/safe or not? (Use ?Y=300 MPa, n=2 - FOS for pin/bearing and ?Allow=300 MPa for bearing).The cross sectional area of the bar is 5cm^2. Use E = 70GPa for aluminum, and E = 200 GPa for steel What is the elongation in segment AB (Aluminum)? a. 7.8571 x10^-3 m b. 6.8571 x10^-3 m c. -7.8571 x10^-3 m d. -6.8571 x10^-3 m What is the elongation in segment BC (Steel)? a. -3x10^-3 m b. 3x10^-3 m c. 2 x10^-3 m d. -2 x10^-3 m What is the elongation in segment CD (Aluminum)? a. -6.8571 x10^-3 m b. 6.8571 x10^-3 m c. -2.8571 x10^-3 m d. 2.8571 x10^-3 m What is the total elongation of the bar? a. 5 x10^-3 m b. 6 x10^-3 m c. 4 x10^-3 m d. -6 x10^-3 m -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.- Direction of the assumption of the equilibrium equation must be shown, no direction will be marked wrong.Torsional T is applied to the hollow axis (G=80 GPa). Ewha Womans University, the maximum shear strain is γmax = 640 × 10-6rad Igo, inside andThe outside diameters are 120mm and 150mm, respectively.a. What is the maximum axial strain?b. Maximum axial stress?c. What is the maximum torque (T) meeting the above conditions?