4. A steel bar of 40 mm x 40 mm square cross-section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E = 200 GPa, the elongation of the bar will be:
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- A steel bar of 40 mm x 40 mm square cross section is subjected to an axial tensile load of 200 kN. If the length of the bar is 2m and E=200 GPa, the elongation of the bar will be:Compute the total elongation caused by an axial load of 147 kN applied to a square bar, tapering from a width of 120 mm to 40 mm in a length of 10 m as shown. Assume E = 202 GPa. Answer must be in mm. 120 mm 40 mm Area = A %D P kN P kN B L = 10 mA steel bar of 40 mm x 40 mm square cross section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E = 200 GPa, the contraction of the bar will be, (a) 1.25 mm, (b) 2.70 mm (c) 4.05 mm (d) 5 40 mm
- A steel bar 40 mm x 40 mm square cross-section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E= 200 GPa, the elongation of the bar will be (a) 1.25 mm (b) 2.70 mm (c) 4.05 mm (d) 5.40 mm e) none of theseA steel rod of 30 mm diameter is enclosed centrally in a hollow copper tube of external diameter 40 mm and 5 mm thick. The composite bar is then subjected to an axial pull of 45 kN. If the length of the compound rod is 150 mm , and the elasticity moduli are Est = 200 GPa and Ecu = 100 GPa , determine: 1.2.2 The load carried by the rod and that carried by the tubeThe 4 mm diameter cable BC is made of a steel with E = 200 GPa. If it is known that the maximum stress in the cable should not exceed 190 MPa and that the elongation of the cable should not exceed 6 mm, find the maximum load P that can be applied as shown in the figure.
- A bar of mild steel 200 mm long and 50 mm × 50 mm in cross section is subjected to an axial load of 200 kN. If E is 200 GPa, the elongation of the bar will beA timber column, 8in. by 8in. in cross section is reinforced on all four sides by steel plates, each plate being 8in. wide and "t"in. thick. Determine the smallest value of "t" for which the column can support an axial load of 300 kips if the working stresses are 1200psi for timber and 20ksi for steel. The moduli of elasticity are 1.5x10^6psi for timber and 29x10^6psi foe steel -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.Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.50 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.85in. Aluminum rod (3) has a diameter of d3 = 1.00in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. Assume a = 2.5 ft, b = 1.5 ft, L1 = 6 ft, L2 = 8 ft, and L3 = 3 ft. (A) Calculate the cross-sectional areas of the three rods. in in.2 (B) For a factor of safety of 2.1, calculate the allowable stresses in the bronze and the aluminum rods. IN KSI. (C) What is the magnitude of load P that can safely be applied to the structure for a minimum factor of safety of 2.1? in kips (D) The pin used at B has an ultimate shear strength of 58 ksi. If a factor of safety of 2.5 is required, determine the allowable shear stress in this pin.…
- Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.40 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.70in. Aluminum rod (3) has a diameter of d3 = 1.00in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. Assume a = 2.5 ft, b = 1.5 ft, L1 = 6 ft, L2 = 8 ft, and L3 = 3 ft.Two 300 mm x 120 mm I-section joists are united by 12 mm thick plates as shown in Figure 2 to form a 7 m long stanchion pinned at both ends. Given a factor of safety of 3, a compressive yield stress of 300 MN/m2 and a Rankine constant a= 1/7500, determine the allowable load which can be carried by the stanchion according to the Rankine-Gordon formulae.A straight and piecewise uniform bar ABCD of total length L is in Case 1 fixed at A in case 1 and in Case 2 fixed at both ends A and D. AB = BC = CD = L/3. The cross section area of segment CD is half of the area of the segment ABC, i.e. AABC = 2ACB. A point load of P is applied at B. Assume that bar material is linear and elastic and the modulus of elasticity is E.