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CE 315- Design of Concrete Structure I Instructor: Dr. E. R. Latifee Page | 12 Analysis of a Beam-Finding Nominal Strength or Capacity of a Beam in Flexure by Equivalent Rectangular Stress

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Page 1: CE 315- Design of Concrete Structure I Instructor: Dr. E ...drlatifee.weebly.com/uploads/4/8/3/2/48322241/analysis_of_beam...Flexure Equations 0.85fc'ab solving for a, 0.85fc'b actual

CE 315- Design of Concrete Structure – I Instructor: Dr. E. R. Latifee

Page | 12

Analysis of a Beam-Finding Nominal Strength or Capacity of a Beam in Flexure by Equivalent Rectangular Stress

Page 2: CE 315- Design of Concrete Structure I Instructor: Dr. E ...drlatifee.weebly.com/uploads/4/8/3/2/48322241/analysis_of_beam...Flexure Equations 0.85fc'ab solving for a, 0.85fc'b actual

CE 315- Design of Concrete Structure – I Instructor: Dr. E. R. Latifee

Page | 13

Example 1 Determine Mn, the nominal or theoretical ultimate moment strength of the

beam section shown in the Figure below

Example 2 (done by Mr. Naim Hassan, AUST student ID- 13-02-03-048) A simply supported beam of 20’’ total depth & width 12” , here 72.5 grade steel (500 W) is used and compressive strength of concrete is 3 ksi. 3#9 bar is used Find the moment capacity of the beam. f y =

60 ksi, f’ c= 3 ksi SOLUTION: GIVEN

h = 20”, d=20”-2.5”=17.5” Now, T=C

AS f y=0.85f’c a b

a=𝐴𝑠𝑓𝑦

0.85𝑓"𝑐 𝑏 =

3×1×72.5

0.85×3×12 ; a=7.11 inch

M n= T (d-a/2) = As f y (d-a/2)

M n = 3 x1x 72.5 x(17.5 - 7.11

2)

M n = 253 k-ft

Page 3: CE 315- Design of Concrete Structure I Instructor: Dr. E ...drlatifee.weebly.com/uploads/4/8/3/2/48322241/analysis_of_beam...Flexure Equations 0.85fc'ab solving for a, 0.85fc'b actual

CE 315- Design of Concrete Structure – I Instructor: Dr. E. R. Latifee

Page | 14

Example 3

Example 4