Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition
Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition
5th Edition
ISBN: 9781259820960
Author: Leet, Kenneth
Publisher: McGraw-Hill Education
Question
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Chapter 8, Problem 17P
To determine

Find the vertical and horizontal deflection of the hinge at C.

Expert Solution & Answer
Check Mark

Answer to Problem 17P

The vertical deflection of the hinge at C is 8.6in._.

The horizontal deflection of the hinge at C is 15.4in._.

Explanation of Solution

Given information:

A single concentrated load of 60 kips is applied at joint B.

Area of all segments of the arch is A=20in.2.

Moment of inertia I=600in.4 and value of E is 30,000ksi.

Procedure to find the deflection of truss by virtual work method is shown below.

  • For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces in all the members of the truss.
  • For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces in all the member of the truss.
  • Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

  • For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
  • For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Find the length of the AB segment of the arch using given Figure.

LAB=302+38.462=48.78ft

Find the length of the BC segment of the arch using given Figure.

LBC=302+(5038.46)2=32.14ft

Find the angle made by the member AB with respect to the horizontal.

θAB=tan1(38.4630)=52°

Find the angle made by the member BC with respect to the horizontal.

θBC=tan1(5038.4630)=21°

Sketch the arch with the forces produced by the P-system as shown in Figure 1.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 8, Problem 17P , additional homework tip  1

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

MA=0Ey(120)60(30)=0Ey=15kips

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+Ey60=0Ay+1560=0Ay=45kips

Summation of moments about C is equal to 0.

MC=0Ax(50)60(30)+45(60)=0Ax=18kips

Summation of forces along x-direction is equal to 0.

+Fx=0AxEx=0Ex=18kips

Find the FP forces and MP moments of each member as follows:

For AB,

FABP=45sin52°18cos52°=46.55kips

MABP=(45cos52°18sin52°)x=13.52x

For BC,

FBCP=15sin21°18cos21°=11.23kips

MBCP=659.520.52x

For ED,

FEDP=15sin52°18cos52°=22.9kips

MEDP=4.95x

For CD,

FCDP=15sin52°18cos21°=22.18kips

MCDP=241.46+7.51x

Consider a dummy load of 1 kips directed vertically at joint C with the forces FQ.

Sketch the reactions produced at the supports by the vertical dummy load at C as shown in Figure 2.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 8, Problem 17P , additional homework tip  2

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

MA=0Ey(120)1(60)=0Ey=0.5kips

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+Ey1=0Ay+0.51=0Ay=0.5kips

Summation of moments about C is equal to 0.

MC=0Ax(50)0.5(60)=0Ax=0.6kips

Summation of forces along x-direction is equal to 0.

+Fx=0AxEx=0Ex=0.6kips

Find the FQ forces and MQ moments of each member as follows:

For AB,

FABQ=0.763kips

MABQ=0.165x

For BC,

FBCQ=0.739kips

MBCQ=8.05+0.25x

For ED,

FEDQ=0.763kips

MEDQ=0.165x

For CD,

FCDQ=0.739kips

MCDQ=8.05+0.25x

Find the vertical deflection at C (ΔCy) as follows:

(1kips)ΔCy=ΣFQFPLAE+ΣMQMPdxEI(1kips)ΔCy=[1AE[(46.55)(0.763)(48.78×12)+(11.23)(0.739)(32.14×12)+(22.9)(0.763)(48.78×12)+(22.18)(0.739)(32.14×12)]+048.78(13.52x)(0.165x)EIdx+032.14(659.520.52x)(8.05+0.25x)EIdx+048.78(4.95x)(0.165x)EIdx+032.14(241.46+7.51x)(8.05+0.25x)EIdx]ΔCy=40,600AE+90,972×123EI=40,60020×30,000+90,972×12330,000×600

ΔCy=8.6in.=8.6in.

Therefore, the vertical deflection at C is 8.6in._.

Consider a dummy load of 1 kips directed horizontally at joint C with the forces FQ.

Sketch the reactions produced at the supports by the horizontal dummy load at C as shown in Figure 3.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 8, Problem 17P , additional homework tip  3

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

MA=0Ey(120)1(50)=0Ey=0.42kips

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+Ey=0Ay+0.42=0Ay=0.42kips

Summation of moments about C is equal to 0.

MC=0Ax(50)0.42(60)=0Ax=12kips

Summation of forces along x-direction is equal to 0.

+Fx=0AxEx+1=012Ex+1=0Ex=12kips

Find the FQ forces and MQ moments of each member as follows:

For AB,

FABQ=0.42sin52°+0.5cos52°=0.64kips

MABQ=0.1354x

For BC,

FBCQ=0.42sin21°+0.5cos21°=0.62kips

MBCQ=6.610.206x

For ED,

FEDQ=0.42sin52°0.5cos52°=0.64kips

MEDQ=0.1354x

For CD,

FCDQ=0.62kips

MCDQ=6.61+0.206x

Find the horizontal deflection at C (ΔCx) as follows:

(1kips)ΔCx=ΣFQFPLAE+ΣMQMPdxEI(1kips)ΔCx=[1AE[(46.55)(0.64)(48.78×12)+(11.23)(0.62)(32.14×12)+(22.9)(0.64)(48.78×12)+(22.18)(0.62)(32.14×12)]+048.78(13.52x)(0.1354x)EIdx+032.14(659.520.52x)(6.610.206x)EIdx+048.78(4.95x)(0.1354x)EIdx+032.14(241.46+7.51x)(6.610.206x)EIdx]ΔCx=6,240AE+160,512.5×123EI=6,24020×30,000+160,512.5×12330,000×600

ΔCx=15.4in..

Therefore, the horizontal deflection at C is 15.4in._.

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