Question

Cracking Moment (Uncracked Concrete Stage) Problem 1) Problem 2.6 (page 55, McCormac and Brown, gth Ed.) Note: Modify the total depth of the beam from 24" to 26" and the depth of the steel rebars from 21" to 23". Transformed Area Method (Concrete Cracked- Elastic Stresses Stage) Problem 2) Problem 2.13 (page 56, McCormac and Brown, 8th Ed.) Note: Make the following changes: Change fe to 5 ksi use the values and dimensions shown in the figure below 1) 2) normal weight concrete f 5000 psi Grade 60 Steel 32 kip d-30" 2k/ft (including beam weight) h34 6M9 bars ← -10 ft (a) Find the reactions at A (left support) and B (right support). (b) Determine the maximum applied moment. (c) Calculate the cracking moment of the beam and verify that the beam has cracked under the applied loads. (d) Calculate the cracked moment of inertia. Note that you will need to first calculate the concrete modulus of elasticity and modular ratio. Do not use the textbook modular ratio of neg e) Compute the flexural stresses. Compressive stress of concrete at the top of the beam and tensile stress at the centroid of the steel bars. Nominal Strength Analysis (Ultimate Strength Stage) Problem 3) Problem 2.29 (page 58, McCormac and Brown, 8 Ed.) Note: Change the 'c from 4ksi to 4.5ksi Hint: find the centroid of the steel bars.

Answer 1

Assumption:

I am assuming all the loads are factored as the dead loads and live loads are not mentioned

(a) Find out support reactions:

Consider the equilibrium condition of the beam

$\sum$ F_y=0

R_A+R_B=32+2x30=92 kips

$\sum$ M_B=0

R_A(30)-32(20)-2(30x15)=0

R_A=51.33 kips

R_B=40.66 kips

(b) Applied moments:

Separate the beam into two parts a point load beam and UDL and then analyze

M1=Wab/L=32x10x20/30=213.33 ft-kips

M2=wL²/8=2x30²/8=225 ft-kips

M=M1+M2=213.33+225=438.33 ft-kips

(c) Cracking moment:

M_cr=f_rI_g/y_t

f_r=7.5x1.0x(5000)^0.5

f_r=530.5 psi

I_g=bh³/12=16x34³/12=52405.333 in⁴

M_cr=530.5x52405.33/17

M_cr=136.27 ft-kips<M_a

The cracking moment is less than the applied moment, so the beam has cracked under applied loads

(d) Cracked moment of inertia:

Calculate n value:

E_c=57000(5000)^0.5=4030.5 ksi

E_s=29000 ksi

A_st=6 in²

n=E_s/E_c=7.19 (consider as 7)

Using the transformed area method, calculate the value of the cracked moment of inertia, let us assume the value of neutral axis from the top as x

bx(x/2)=nd^'(d-x)

16x(x/2)=7x4x(30-x)

8x²=28(30-x)

8x²+28x-840=0

x=8.64 in

I=bx³/3+nA_st(d-x)²

I=16x8.64³/3+7x6x(30-8.64)²

I=22601.85 in⁴

(e) The stresses:

From the above values:

f_c=My/I=12x433.8x1000x8.64/(22601.85)=1990 psi (Here the value of y is 8.64 in obtained from above)

f_s=nMy/I=7x12x433.8x1000x(21.36)/22601.85=34437 psi (Here the value of y is d-x=30-8.64=21.36 in )