If the wood used for the T beam has an allowable tensile and compressive stress of allowt= 4 MPa and allow = 6 MPa respectively, determine the maximum allowable internal moment M that can be applied to the beam. 100 mm 100 mm 50 mm 50 mm 200 mm 100 mm 100 mm M A 50 mm 200 mm 60% B 50 mm
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- Topic: COMBINED STRESS-AXIAL TENSION AND FLEXURE BENDING: STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee Questions A Tension member with no holes is subjected to axial loads of PD=68kN & PL=64kN.It is also subjected with bending moments of MDy=40kN-m & MLy=55kN-m. Is themember adequate? Steel is of A992 Gr 50 Specs. Use LRFD. Neglect the weight of the beam. Section properties:Lp = 1.863 md=459.99mmtw=9.02mmzx= 1835x10^3 mm^4Lr = 5.305 mLb = 4.8 mbf=191.26mmtf=16mmSx=1611x10^3 mm^4 Sy=195x10^3 mm^4Zy=303x 10^3 mm^4 Cb=1.32In the connection shown, 1/4-in. side and end fillet welds are used to connect the 3-in.-by-1 in. tension member to the plate. The applied load is 60,000 lb. Find the required dimension L. The steel is ASTM A36 and the electrode used is an E70. - Fillet welds L 3" Activate Windows Go to Settings to activate Wir TABLE 19.2 Allowable stresses in ksi (MPa) 12: 12/1 DELL end F12 insert prt sc F10 home F11 F6 F8 F9 F3Calculate the shear yield stress (ksi) of A588 steel.
- Situation No. 2 A bronze bar is fastened between a steel bar and an aluminum bar as shown below. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the assembly is suitably braced to prevent buckling. Use Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa. Bronze Steel A = 480 mm? Aluminum A = 650 mm A = 320 mm 3P 4P 2P 1.0 m 2.0 m 1.5 mRequired information A square yellow-brass bar must not stretch more than 2.4 mm when it is subjected to a tensile load of 34 kN. Given: E= 105 GPa and that the allowable tensile strength is 104 MPa, determine (a) The allowable side length of the cross-section (b) is the allowable value a maximum or a minimum? Why? (c) The allowable length of the bar (d) is the allowable length a maximum or a minimum? (a&b) The allowable side length, for the square cross section, is given as [ (mm) and the value is a (Click to select)Specify a suitable aluminum casting alloy for a bar with square crosssection. Length of one side of the cross-section is 50 mm. The bar is subjected toan impact loading with a tensile force of 60 kN
- Q Search MT ACTIVITY 2M Close Activity File Home Insert Draw View Help Open in Desktop App O Tell me what you want to do v 11 ev Av A A Styles v E Tags v abc v Calibri ... SW The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F. Assume a = 6.5 × 10- Teams Assignments 国 Calendar Calls 6 in/(in-°F) for steel and 12.8 x 10-6 in/(in-°F) for aluminum. Files Aluminum A = 2 in? E = 10 x 10° psi Steel A = 3 in? E = 29 x 10° psi ... 15 in 10 in 田 Help1. Three metal strips, each 40 mm height, are bonded together to form the composite beam shown. The modulus of elasticity is 210 GPa for the steel, 105 GPa for the brass, and 70 GPa for the aluminium. If the allowable bending stress for the aluminum (Gallow)al= 100 MPa, for the steel (Gallow)st 150 MPa and (Gallow)br=200 MPa for brass determine the maximum allowable intensit of w of the uniform distributed load. 2w 2m Aluminum Brass Steel 40 mm- 10 mm 10 mm 20 mmTwo T-shape sections are welded together to create a composite l-beam section as shown below. Determine the safely uniform load w (kN/m) that the beam can carry throughout its entire span. 270mm 30mm 20mm 160mm Welded Joint 130mm 30mm 135mm Beam Length = 4 m Consider the following stresses: fb(T) 3 125 МPа fb(C) = 130 MPa x at the welded joint = 20 MPa Find: 1. The Neutral Axis from the base 2. moment of inertia with respect to the horizontal centroidal axis 3. Q at the welded joint 4. Considering the maximum flexural stress in tension, WNAX is equal to KN/m. 5. Considering the maximum flexural stress in compression, WMAX, is equal to 6. Considering the stress at the welded joint, wMex is equal to kN/m.
- Determine the stress and total load on each stringers if the web is negligible. Estainless steel = 28000 ksi, Esteel = 29000 ksi, EMagnesium = 6500 ksi, EAlum. Ally = 10500 ksi 6" Steel 10 .10 a Stainless Steel My = 5000"# 10" Mx = 10000"# Magnesium .1 Alloy Area of Each 10 Alum. Alloy Stringer = 1 sq. in. 4"く> Q Search MT ACTIVITY 2M Close Activity File Home Insert Draw View Help Open in Desktop App O Tell me what you want to do v11 I U クv Av S A。 ニvニ A Styles v E Tags v abc Calibri ... Teams SW Assignments The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F. Assume a = 6.5 × 10 6 in/(in-°F) for steel and 12.8 × 10-6 in/(in·°F) for aluminum. 国 Calendar Calls Files ... Aluminum A = 2 in? E = 10 x 10° psi Steel A = 3 in? E = 29 x 10° psi 田 15 in 10 in Help2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 25 mm= DIAMETER OF BOLTS 45 45mm fmmy C. 90mm 90mm 90mm Comm P