For the truss shown, the working stresses are 20 000 psi in tension and 12 000 psi in compression. (A reduced stress in compression is specified to reduce the danger of buckling.) 12 ft Determine the smallest allowable 8 ft cross-sectional area of member BD of the truss shown. 8 ft 8 ft 8 ft 36 kips FIGURE 8 Determine the smallest allowable cross-sectional area of member BE of the truss shown. Determine the smallest allowable cross-sectional area of member CE of the truss shown.

For the truss shown, the working stresses are 20 000 psi in tension and 12 000 psi in compression. (A reduced stress in compression is specified to reduce the danger of buckling.) 12 ft Determine the smallest allowable 8 ft cross-sectional area of member BD of the truss shown. 8 ft 8 ft 8 ft 36 kips FIGURE 8 Determine the smallest allowable cross-sectional area of member BE of the truss shown. Determine the smallest allowable cross-sectional area of member CE of the truss shown.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

A.) If the F = { - 200i + 400j } N , determine the force in each member and state if its in compressions or tension. A.) Using the same given, if the members have a square cross-section with a diagonal of 11.31cm and thickness of 1cm, determine the axial stressess and the factors of safety on each of the member, if the maximum allowable tensile strength is 60kPa and the maximum allowable compressive strength is 70kPa NOTE: Complete solution and units, free body diagrams, table data if possible <up voting!>
A six-pack is shaken so that the pressurein each can is p = 20 psi. A board is placed onthe six pack and a student(s) weighing Wstands on the board so his weight is evenlydistributed to the six cans. The aluminumproperties are E = 10,000 ksi, Sy = 35 ksi, andν = 0.33. Each can has an average radius ofR = 1.25 in., thickness t = 0.01 in., and lengthL = 4.50 in.(a) Determine the critical weight of thestudent, Wcr , so that a stress element on thesurface of the can is in a state of uniaxial stress.Assume that no support is provided by thecontained liquid and the cans do not fail bybuckling (being crushed). (b) Determine thestrain in the longitudinal direction of the candue to pressure p = 20 psi and force Wcr.(c) Determine the strain in the hoop directionof the can due to pressure p = 20 psi and forceWcr
The rigid bar of negligible mass, pinned at one end, is supported by a steel rod and aluminum rod asshown. What maximum load P can be applied without exceeding stress in the steel of 130MN/m2or in the aluminum of 90MN/m2
1. By method of sections, determine the force in member CD, CF, and FG of the Warren truss. Solve for the forces, axial stresses and the factors of safety if the member is a solid-circular cross-section that has a diameter of 30 mm and a maximum allowable stress of 10kN/m^2. (Include free body diagram)
The rigid bar ABCD of negligible weight is initially horizontal, and the steel rods attached at A and C are stress-free. If the temperature is dropped to −40°F and the 20-kip load is then applied, determine the stress in each steel rod at A and C Use E=29x106 psi and ∝=6.5x10−6/° F for steel.
Three steel rods (E = 200 GPa) supports a 50 kN load P. Each of the rods AB and CD has a 300 mm2 cross-sectional area and rod EF has a 600 mm2 cross-sectional area.State whether the structure is statically determinate or indeterminate. Why?Determine the force and the stress in each rod.Determine the deformation of each rod. Note: The link BED is rigid, and BE = ED.
A 75 mm diameter compound bar is constructed by shrinking a circular brass bush onto the outside of a 50 mm diameter solid steel rod. If the compound bar is then subjected to an axial compressive load of 160 kN determine the load carried by the steel rod and the brass bush and the compressive stress set up in each material. For steel, E = 210 GN/m2; for brass, £ = 100 GN/m2
All members of the truss shown have a cross-sectional area of 500mm2 and all pins have a diameter of 20mm2. Determine: (a)The axial stresses in members BC and DE. (b)The shear stresses in the pin at A, assuming the pin is in double shear.
Two steel tubes are joined at B byfour pins (dp = 11 mm), as shownin the cross-section a-a in thefigure. The outer diameters of thetubes are dAB = 40 mm and dBC = 28mm. The wall thicknesses are tAB =6 mm and tBC = 7 mm. The yieldstress in tension for the steel is y= 200 MPa and the ultimate stressin tension is U = 340 MPa. Thecorresponding ultimate value inshear of the pin is 140 MPa.Assume that the factors of safetywith respect to the yield andultimate stress are 4 and 5,respectively.a.) Calculate the allowable tensileforce Pallow considering tension inthe tubes.b.) Recompute Pallow for shear inthe pins.
Rod BD is made of steel (E = 29 × 106 psi) and is used to brace the axially compressed member ABC. The maximum force that can be developed in member BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the smallest-diameter rod that can be used for member BD. Take P = 148 kips.The smallest-diameter rod that can be used for member BD is in.
The horizontal bar shown is pinned at O and is supported with a helical springat L1 from O. L1=2m, L2=1m. The spring has 30 turns and the diameter of the wire usedis 20mm. The diameter of the spring is 125mm. Modulus of Rigidity = 83GPa. Determine the shear stress of the helical spring loaded as shown if w = 25kN/m.
Two forkend pieces are to be joined together by a single steel pin of 25mmdiameter and they are required to transmit 50 kN. Determine the minimum crosssectional area of material required in one branch of either fork if the stress in the forkmaterial is not to exceed 180 MN/m2. What will be the maximum shear stress in thepin?