Question

A beam with a cross-section of 0.3 m x 0.7 m has an effective depth of 0.638 m and a span of 4.30 m. Use the following data below to determine (a) the steel requirement due to flexure, (b) ultimate threshold torsion, and (c) spacing of stirrup due to combined shear and torsion:

M_u = 322 KN-m

V_u = 317 KN

T_u = 42 KN-m

f'c =28 MPa

f_yt = 414 MPa

12 mm stirrup diameter

cover of 0.04 m

Answer 1

Since it is not mentioned as to which Code to use for this question, i am solving the question based on the Indian Standard Code (IS 456:2000)

a) Steel requirement (A_{​​​​​​st}​​​​)

Since M_{​​​​​​u} =322KN-m

M_{​​​​​​u} = 0.87f_yA_st×(d-0.42x_u) ...... (1)

(tensile force×LeverArm)

X_{​​​​​​u} = depth of Neutral Axis for fyt = (0.87f_y A_{​​​​​​}​​​​​​_​​​st )/(.36fc b)

b= width =0.3m =300mm

Putting values for Xu we get = 0.119Ast

d= effective depth=0.638m = 638mm

Now putting the respective values in the eq (1)

322×10⁶ N-mm = 0.87×414×A_st × (638- 0.42×0.119Ast)

Solving we get,

A_{​​​​​​st} = 1602.39 mm2

b) Ultimate threshold torsion

M_{​​​​​​T} = T_{​​​​​​u}/1.7 × (1+D/b)

D=total depth =0.7m =700mm

b= width=0.3m =300mm

T_{​​​​​​u} = 42Kn-m =42×10⁶ N-mm

Putting values:

M_{​​​​​​T} =82.35KN-m

c) Stirrup Spacing due to combined action of Shear and Torsion

Asv = (T_{​​​​​​u} Sv)/bd (0.87f_y ) + (V_{​​​​​​u} Sv)/2.5d(0.87f_y ) ..(2)

Where,

Sv = Spacing of stirrups due to combined action of Shear and Torsion

Asv = Area of stirrups provided (here we provide double legged stirrup) dia given 12 mm so area =2×(π/4 ×12²)

V_{​​​​​​u} =317 KN =317000N

T_{​​​​​​u} =42KN-m or 42000000N-mm

Putting respective values in eq (2) and solving for the only unknown value Sv

We get,

Sv=194.72mm

Provide 190mm.