Challenge 2

1. solution :given loads : dead load=100KN

super imposed load:50KN

live load: 200KN

wind load:+150KN\ -150KN

earthquake load or Quasistatic load:+200KN\-200KN.

The design load on structures as per working stress method is eqal to

=DL+LL+QL+WL

=100KN+200KN+150KN+200KN

=650KN.

or other combination. =100KN+200KN-150KN-200KN

=-50KN.

the greatest among should be considered. so, the design load as per working stess method will be 650KN.

The design load on structures as per limit state method is

1. = 1.5(DL+LL)=450KN

2. =1.5(DL+QL)=450kN

=1.5(100-200)=-150KN

3. =1.2(DL+LL+QL)=1.2(100+200+200)=600KN

=1.2(100+200-200)=120KN.

so, the greatest among 3 results should be considered, the design load will be 600KN.

2. solution.

Given: D=150mm,

ϕ=8mm.

S=125mm.

c=20mm.

grade of concrete=M30.

grade of steel=Fe-415.

Effective depth d= D-C-ϕ\2 =d

=150-20-4=126mm.

area of one steel bar,Ast=π\4 xdsqare

no of bars in 1m width=(1000\S)

so, Ast=(1000\S)x(π\4)x(ϕ)2

=128π mm2.

Now, we have to classify the section as Under reinforced, Over reinforced Or balanced section, so.

if . Xu > Xu max it is over reinforced

if Xu = Xu max it is balanced section

and ,if Xu <Xu max it is under-reinforced section.

so, calculating Xu.

Xu= (o.87fyAst\0.36fckb)

given, fy=415, and fck=30.

b= width of slab=1000mm.

so, Xu=13.443mm.

now , we have to calculate Xu max.

Xu max \d =0.0035\0.0055+(0.87fy\Es).

Xumax\d=0.4791.

Xumax=0.4791x126mm=60.366.

so, Xu<Xu max so, it is under reinforced section.

so, The moment of resistance for under reinforced section equals to.

MR= T X Z

= 0.87 Fy Ast (d-0.42Xu)

=0.87X 415x128πX(126-0.42X13.443)

=17.474 KN-m.