What is the maximum spacing by code not by strength for stirrups in a RC beam if the shear strength required by the steel is more than twice the strength of the concrete? 'd' for the beam is 20 inches. 10" 24" 12" 5"

Combined Stresses Given Hollow Steel Section (HSS) that is 4"x6" and 1/2" thick, with axial and transverse loading shown, determine the total normal stress at a. The 5k load on the bottom of the beam b. At the mid-span in the middle of the beam C. At the 10k load on top of the beam 5k 10k A 5' 10' 5' B CENTROID 4" 5 k 3k 8k 6" 2' 6.25k 8.75k +6.25k SECTION +1.25k 'V" -8.75k +43.75k-ft. +31.25k-ft. 'M' 8k 5 k + tension P = Ok %3D

30" Design the reinforcement for the T-beam in the middle which carries a live load of 200 lb/ft². The length of the beam is 25 ft. In your design assume concrete with a compressive strength of 3,000 psi density 150 lb/ft³ and Grade 60 Steel. 27" 12" 10'-0" 12" 10'-0" 12"

+ „blackboard.com/ultra/courses/_96287_1/cl/outline Question Completion Status: 4 8 9 10 A two span beam subjected to shear and flexure only is reinforced as follows: SECTION @ MIDSPAN @ FACE OF SUPPORTS TOP BARS 2-020 mm 5-Ø20 mm BOTTOM BARS 3-020 mm 2-020 mm Given: Stirrup diameter, ds = 10 mm Concrete fc = 21 MPa Steel rebar = 415 MPa Stirrup fy = 275 MPa %3D Beam size b xh = 270 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = (in 5 decimal places) kN m (nearest whole Design Moment strength of section at midspan for positive bending = 32°C Light rain ^ O O G ) ENG 1:42 pm ロ