Lecture 5, Beam design

CE 318

STRUCTURE ANALYSIS AND DESIGN II LAB

LECTURE 5 BEAM DESIGN

SEMESTER: SPRING 2021

COURSE TEACHER: SAURAV BARUA CONTACT NO: +8801715334075

EMAIL: saurav.ce@diu.edu.bd

LECTURE PLAN

Beam flexural design

Beam shear design

Reinforcement detailing of beam

COMBINATION OF MOMENT (k-ft/ft)

Load combo

A B C D

Envelop -125.29 143.13 -286.54 206.06

Effective depth

• Take maximum moment and check d

d = √(286.54/(0.9 x 0.016 x 60 x 1x (1-0.59x0.016x60/3))

= 20.22”

d_req = 20.22”

Assume, beam size = 12”x 25” (Explanation in class!!) d_provided = 25-2.5 = 22.5” > d_req ok

REINFORCEMENT

• Reinforcement at B

A_s = 143.13x 12 /0.9 x 60 x (22.5-1/2), assume a =1”

= 1.45 in²

a = 1.45 x 60 / 0.85 x 3 x 12 = 2.84 A_s = 1.51 in²

So, provide Ø 20mm nos. = 1.51/ 0.44 ≈ 4 Use 5nos. Ø20mm

REINFORCEMENT

• Reinforcement at D

A_s = 206.06x 12 /0.9 x 60 x (22.5-2.84/2), assume a =1”

= 2.17 in²

So, provide Ø 20mm nos. = 2.17/ 0.31 ≈ 5 Use 5nos. Ø20mm

• Reinforcement at A

A_s = 125.29x 12 /0.9 x 60 x (22.5-2.84/2), assume a =1”

= 1.32 in²

So, provide Ø 16mm nos. = 1.32/ 0.31 ≈ 5

Use 3nos. Ø20mm + 2nos. Ø16mm (Extra top) [explain in class!!]

REINFORCEMENT

• Reinforcement at C

A_s = 286.54x 12 /0.9 x 60 x (22.5-2.84/2), assume a =1”

= 3.02 in²

So, provide 3nos. Ø20mm = 1.32 in²

Extra top = 3.02-1.32 = 1.7 in², Ø25mm nos. = 1.7/0.61 = 3 Use 3nos. Ø20mm + 3nos. Ø25mm (Extra top)

COMBINATION OF SHEAR (k/ft)

Load

combo A B C D

Envelop 70.11 8.87 -80.14 10.66

Shear reinforcement

Shear at C

V_U = 80.14 k/ft

ØV_c = 2 Ø√f_c^’b_w d = 2 x 0.85 x √3000 x 12 x 22.5

= 25141 lb/1000 = 25.1kip

Shear taken by steel, Vs = V_u/Ø – V_c = 80.14 – 25.1/0.85 = 50.6 kip 4 Ø√f_c^’b_w d = 50.2kip > V_s

• Use ø10mm bar

S_req = A_v f_yd / V_s = 0.22 x 60 x 22.5/ 50.6 ≈ 5.5” c/c Provide ø10mm bar @ 5.5” c/c stirrups

Upto l/4 = 18/4 = 4’-6” distance Also same stirrups for A.

Shear at D

V_U = 10.66 k/ft

Shear taken by steel, V_s = V_u/Ø – V_c = 10.66 – 25.1/0.85 < 0 4 Ø√f_c^’b_w d = 50.2kip < V_s

So, Provide minimum reinforcement

• S_max = A_v f_y/ 50 b_w = 0.22 x 60000/ 50 x 12 = 22”

or d/2 = 22.5/2 = 11.25”

or 24” , whichever is smaller Provide Ø10mm @ 11” c/c Same for B also.

B1 BEAM LONG SECTION

• a = 2nos. Ø16mm extra top, b = 3nos. Ø20mm, c = 3nos. Ø25mm extra top, d= 5nos. Ø20mm extra top. exterior l/5 = 3’ and interior l/4= 4’-6” extra top

• e = Ø10mm stirrups @ 5.5”c/c, f = Ø10mm stirrups @ 11”c/c

• a = 2nos. Ø16mm extra top, b = 3nos. Ø20mm, c = 3nos. Ø25mm extra top, d= 5nos. Ø20mm extra top. exterior l/5 = 3’ and interior l/4= 4’-6” extra top

• e = Ø10mm stirrups @ 5.5”c/c, f = Ø10mm stirrups @ 11”c/c