Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 6, Problem 6.3.16P

A reinforced concrete slab (see figure) is reinforced with 13-mm bars spaced 160 mm apart at d = 105 mm from the top of the slab. The modulus of elasticity for the concrete is Ec= 25 GPa, while that of the steel is £s = 200 G Pa. Assume that allowable stresses for concrete and steel arecrac = 9.2 MPa and u s = 135 MPa.

l()5 mm

  1. Find the maximum permissible positive bending moment for a l-m wide strip of the slab.

What is the required area of steel reinforcement, A^ if a balanced condition must be achieved? What is the allowable positive bending moment? (Recall that in a balanced design, both steel and concrete reach allowable stress values simultaneously under the design moment.)

  Chapter 6, Problem 6.3.16P, A reinforced concrete slab (see figure) is reinforced with 13-mm bars spaced 160 mm apart at d = 105

(a)

Expert Solution
Check Mark
To determine

The maximum permissible positive bending and balanced condition for both steel and concrete.

Answer to Problem 6.3.16P

The maximum permissible positive pending moment is10.193KN.m.

Explanation of Solution

Given: .

  Es=200GP

  Ec =25GPa

D =0.13m

Calculation: .

The modular ratio of the momentum is,

  n=EsEc.

  n=20025n=8.

The number of reinforcement bar provided is,

  N=bs.

Substitute b=100mm, s=160mm

  N=1000160N=6.

The molecular ratio of the moment isn=8..

The area of reinforcement bar is stated as,

  As=N×π×D24As=6×π×0.01324As=7.964×104m2.

Moment of area of concrete section = Moment of area of steel about neutral Axis..

  by×y2=nAs(dy)12.by2=nAs(dy)by22+nAsynAd=0 ..

Using Quadratic formula,

  x=q±(q)2(4pr)2p.

Substitute p=b2, q=nAsr=nAsd.

  y=nAs±(nAs)2+(4×b2×nAsd)2×b2.

  y=(8×7.964×104)±(8×7.964×104)+(4×12×8×7.964×104×0.105)2×12y=(6.3712×103)+0.03711y=0.0308m.

The location of neutral axis for the top layer, y =0.0308m.

Moment of inertia of the transformed section is,

  IT=13by3+nAS(dy)2.

  IT=13×1×(0.0308)3+8×7.964×104(0.1050.0308)2IT=4.4817×105m4.

The bending moment at the top layer of concrete..

  Mc=σacyIT.

Substitute y=0.0308m,IT =4.487×105.

σac ,=(9.2×106).

  Mc=(9.2×106)0.0308(4.487×105).

  Mc=13.41KN.m.

The bending moment at the top layer of concrete..

  Ms=σasn(dy)IT.

Substitute the value

  Ms=(135×106)8×(0.1050.0308)(4.487×105).

  Ms=10.193KN.m.

Conclusion: .

The maximum permissible positive pending moment is10.193KN.m.

(b)

Expert Solution
Check Mark
To determine

To find: The allowable value of moment in steel in positive pending.

Answer to Problem 6.3.16P

The allowable value of moment in steel in positive pending is15.77KN.m.

Explanation of Solution

Calculation: .

Substitute y equation above

  b×nAs±(nAs)2+(4×b2×nAsd)2×b2×σac2=As×σas.

  b×8As±(8As)2+(4×12×8Asd)b×σac2=As×σas.

Substitute the value n=8,σac =9.2×106 , b=1m,d=0.105m,σas =135×106..

  8As±(8As)2+(4×12×8As×0.105)×9.2×1062=As×135×10.

  8As±(8As)2+(1.68As)×9.2×1062=As×135×106.

  [±(8As)2+(1.68As)]=37.3478As.

Take squaring on both sides,

  (8As)2+(1.68As)=37.34782As2.

  (1,330.8582As)2(1.68As)=0As=1.2623×103m2.

Required area of steel reinforcementAs=1.2623×103m2..

Moment of area of concrete section = Moment of area of steel about neutral Axis..

  by×y2=nAs(dy)12.by2=nAs(dy)by22+nAsynAd=0.

Use quadratic equation..

  x=q±(q)2(4pr)2p.

Substitute p=b2, q=nAsr=nAsd..

  y=nAs±(nAs)2+(4×b2×nAsd)2×b2.

  y=.

  (8×1.2623×103)±(8×1.2623×103)+(4×12×8×1.2623×103×0.105)2×12.

  y=(0.01)+0.04711.

  y=0.0371m.

The allowable positive pending moment in concrete is15.77KN.m..

  IT=13×1×(0.0371)3+8×1.2623×103(0.1050.0371)2IT=6.3579×105m4.

The allowable positive pending moment in concrete is,

  Mc=σacyIT.

Substitute the value y=0.0371m,σac =9.2×106 ,IT =(6.3579×105).

  Mc=(9.2×106)0.0371(6.3579×105)Mc=15.77KN.m.

The allowable positive pending moment in steel is,

  Mc=σasnyIT.

  Mc=(135×106)8×0.0371(6.3579×105)Mc=28.92KN.m.

Conclusion: .

The lesser value of moment in concrete and steel as allowable value of moment..

Allowable positive bending moment in steel is15.77KN.m.

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Chapter 6 Solutions

Mechanics of Materials (MindTap Course List)

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