Bundle: Steel Design, Loose-leaf Version, 6th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
6th Edition
ISBN: 9781337761505
Author: William T. Segui
Publisher: Cengage Learning
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Question
Chapter 9, Problem 9.5.2P
To determine
(a)
The adequate strength of the composite section by LFRD method.
To determine
(b)
The adequate strength of the composite section by ASD method.
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06.
The composite bar, firmly attached to unyielding supports, is initially stress-free. What maximum axial load P can
be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel?
Steel
Aluminum
A = 1.25 in.?
E = 10 x 106 psi
A = 2.0 in.2
E = 29 x 106 psi
SUMMARY OF ANSWERS
P
kips
-15 in:
-12 in.-
CS Scanned with CamScanner
A simply supported beam is 25 × 50 cm deep and has 2-20 mm Fe415 grade steel bars going into the
support shown in figure below. If the shear force at the centre of support is 110 kN at service loads.
Assume M20 mix, design bond stress (ta) = 1.2 MPa.
Take clear cover to steel = 25 mm
2-20 mmo
with referecne to the information provided,
(a) The required development length
(b) The anchorage length
00 mm
A simply supported beam is reinforced with 5-ϕ25 mm at the bottom and 2-ϕ20 mm at the top of the beam. Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc’= 28 MPa, fy = 415 MPa. Assume bars laid out in single layer.
Calculate the following if the limiting tensile steel strain is 0.004 for a ductile failure:
Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = __
mm
Chapter 9 Solutions
Bundle: Steel Design, Loose-leaf Version, 6th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
Ch. 9 - Prob. 9.1.1PCh. 9 - Prob. 9.1.2PCh. 9 - Prob. 9.1.3PCh. 9 - Prob. 9.1.4PCh. 9 - Prob. 9.1.5PCh. 9 - Prob. 9.1.6PCh. 9 - A W1422 acts compositely with a 4-inch-thick floor...Ch. 9 - Prob. 9.2.2PCh. 9 - Prob. 9.3.1PCh. 9 - Prob. 9.3.2P
Ch. 9 - Prob. 9.4.1PCh. 9 - Prob. 9.4.2PCh. 9 - Prob. 9.4.3PCh. 9 - Prob. 9.4.4PCh. 9 - Prob. 9.4.5PCh. 9 - Prob. 9.5.1PCh. 9 - Prob. 9.5.2PCh. 9 - Prob. 9.5.3PCh. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Prob. 9.7.1PCh. 9 - Prob. 9.7.2PCh. 9 - Prob. 9.7.3PCh. 9 - Prob. 9.7.4PCh. 9 - Prob. 9.8.1PCh. 9 - Prob. 9.8.2PCh. 9 - A beam must be designed to the following...Ch. 9 - Prob. 9.8.4PCh. 9 - Prob. 9.8.5PCh. 9 - Prob. 9.8.6PCh. 9 - Prob. 9.8.7PCh. 9 - Prob. 9.8.8PCh. 9 - Use the composite beam tables and select a W-shape...Ch. 9 - Prob. 9.8.10PCh. 9 - Prob. 9.10.1PCh. 9 - Prob. 9.10.2P
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- A simply supported beam is reinforced with 5-ϕ25 mm at the bottom and 2-ϕ20 mm at the top of the beam. Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc’= 28 MPa, fy = 415 MPa. Assume bars laid out in single layer. Calculate the following if the limiting tensile steel strain is 0.004 for a ductile failure: Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = mm Design strength of the beam section to the nearest whole number = kN·m Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = kN/m (to the nearest whole number)arrow_forwardA simply supported beam is reinforced with 5-ϕ25 mm at the bottom and 2-ϕ20 mm at the top of the beam. Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc’= 28 MPa, fy = 415 MPa. Assume bars laid out in single layer. Calculate the following if the limiting tensile steel strain is 0.004 for a ductile failure: Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = kN/m (to the nearest whole number)arrow_forwardA simply supported beam is reinforced with 4-ø28 mm at the bottom and 2-ø20 mm at the top of the beam, Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc'=28 MPa, fy=415 MPa. Assume bars laid out in single layer. Calculate the following if the limitin tensile steel strains is 0.004 for a ductile failure: Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = _____________mm Design strength of the beam section to the nearest whole number =____________ kN-m Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = _____________ kN/m (to the nearest whole number)arrow_forward
- PLEASE WRITE YOUR COMPLETE SOLUTION AND BOX THE FINALS ANSWERS. THANK YOU. A composite bar consists of steel segments with different cross-sections are rigidly fastened as shown. Axial loads are applied at the positions indicated. The stress must not exceed 116 MPa in tension, and 77 MPa in compression. Determine the following: (a) the safe value of P to the nearest safe 1 kN, if the diameter of segment 1 is 100 mm; (b) the required side length of the equilateral triangle segment 2 to the nearest safe 25 mm; and (c) the required side length of the regular pentagon segment 3 to the nearest safe 25 mm.arrow_forwardA simply supported beam is reinforced with 5-p25 mm at the bottom and 2-020 mm at the top of the beam. Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc'= 28 MPa, fy = 415 MPa. Assume bars laid out in single layer. Calculate the following if the limiting tensile steel strain is 0.004 for a ductile failure: Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = mm Design strength of the beam section to the nearest whole number = kN -m Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = kN/m (to the nearest whole number)arrow_forward8. The composite bar is stress-free before the axial loads P1 and P2 are applied. Assuming that the walls are rigid, calculate the stress in each material if P1 = 153 kN and P2 = 80 kN. Steel A = 900 mmA = 2000 mm A = 1200 mm2 E - 200 GPa Aluminum Bronze E - 70 GPa E- 83 GPa 500 mm 250 mm' 350 mmarrow_forward
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- 1 Situation 19. A composite beam shown is subjected to a bending moment of 500 kN-m. Determine the maximum flexural stress; 300 VIEW INNOVATIONS 200 200 100 55. Experienced by the steel 56. Experienced by the aluminum A. 0.03 B. 25.24 57. Experienced by the brass A. 25.24 B. 37.86 REVIEW I EW INNOVATIONS Steel, E=200 GPa Aluminum, E=70 GPa Brass, E-105 GPa wwwww C. 48.2 D. 72.11 C. 16.8 C. 16.8 D. 11.21 VIEW INNOVATIONS REVIEW INNOVATIONS REVIEW ISarrow_forwardtwo span beam subjected to shear and flexure only is reinforced as follows: Section @ midspan @ face supports Top bars 2-20mm 5-20mm Bottom bars 3-20mm 2-20mm Given Stirrup diameter: 10 mm concrete fc = 21 mpa steel rebar fy = 415 mpa stirrup fy = 275 mpa beam size b x h = 270 mm x 450 mm assume all bars laid out in single layer calculate the following: a) tensile steel ratio in positive bending moment at midspan (round off to 5 decimals) b) design moment strength of section at midspan for positive bending ( in kN•m , nearest whole number) c) nominal design strength of section at face of support for negative bending ( in kN•m , nearest whole number)arrow_forward06. The composite bar, firmly attached to unyielding supports, is initially stress-free. What maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel? Steel A = 2.0 in.2 Aluminum A = 1.25 in.2 E = 10 x 10° psi E = 29 x 10° psi SUMMARY OF ANSWERS P kips -15 in- -12 in.-arrow_forward
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