Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
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Chapter 10, Problem 11P
To determine

Find all the reactions of the beam, draw the shear force and bending moment diagram, and determine the deflection at the midspan of segment BC.

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Answer to Problem 11P

The slope at B is 96EI_.

The slope at C is 96EI_.

The end moment of member AB is 48kips-ft_.

The end moment of member BA is 84kips-ft_.

The end moment of member BC is 84kips-ft_.

The end moment of member CB is 84kips-ft_.

The end moment of member CD is 84kips-ft_.

The end moment of member CD is 48kips-ft_.

The deflection under the load is

Explanation of Solution

Apply the sign conventions for calculating reactions 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 positive and the counter clockwise moment as negative.

Determine the fixed end moment of each member as shown below;

FEMAB=wL8=30(16)8=60kips-ftFEMBA=wL8=30(16)8=60kips-ft

FEMCD=wL8=30(16)8=60kips-ftFEMDC=wL8=30(16)8=60kips-ft

FEMBC=wL212=2(24)212=96kips-ftFEMCB=wL212=2(24)212=96kips-ft

Determine the end moments of each member as shown below;

MAB=2EIL(2θA+θB)+FEMAB=2EI16(2(0)+θB)60=EI8(θB)60MBA=2EIL(2θB+θA)+FEMBA=2EI16(2θB+0)+60=EI8(2θB)+60

Express the relationship between slope at C and B as follows:

θC=θB

Determine the end moments of each member as shown below;

MCD=2EIL(2θC+θD)+FEMCD=2EI16(2θB+0)60=EI8(2θB)60MDC=2EIL(2θD+θC)+FEMCD=2EI16(2(0)θB)+60=EI8(θB)+60

MBC=2EIL(2θB+θC)+FEMBC=2EI24(2θBθB)96=E(1.5I)12(θB)96MCB=2EIL(2θC+θB)+FEMCB=2E(1.5I)24(2θB+θB)+96=E(1.5I)12(θB)+96

Apply Equation of equilibrium at joint B;

MBA+MBCEI8(2θB)+60+E(1.5I)12(θB)96=00.25EIθB+0.125EIθB=36θB=96EI

Determine the slope at C using the relation;

θC=θBθC=96EI

Hence, the slope at B is 96EI_.

Hence, the slope at C is 96EI_.

Determine the end moments of each member as shown below;

MAB=EI8(96EI)60=48kips-ftMBA=EI8(2(96EI))+60=84kips-ftMCD=EI8(2(96EI))60=84kips-ftMDC=EI8(96EI)+60=48kips-ft

MBC=E(1.5I)12(96EI)96=84kips-ftMCB=E(1.5)12(96EI)+96=84kips-ft

Hence, the end moment of member AB is 48kips-ft_.

Hence, the end moment of member BA is 84kips-ft_.

Hence, the end moment of member BC is 84kips-ft_.

Hence, the end moment of member CB is 84kips-ft_.

Hence, the end moment of member CD is 84kips-ft_.

Hence, the end moment of member CD is 48kips-ft_.

Show the free body diagram of support A, span AB, support B, span BC, support C, span CD, and support D as in Figure (1).

Fundamentals of Structural Analysis, Chapter 10, Problem 11P , additional homework tip  1

Consider span AB;

Determine the shear VBA using the relation;

Take moment about A;

MA=0VBA×16+(30×8)48+84=0VBA=27516VBA=17.3kips

Determine the shear VAB using the relation;

V=0VAB+VBA30=0VAB=3017.3VAB=12.7kips

Hence, the vertical reaction at A is 12.7kips_.

Consider span BC;

Determine the shear VCB using the relation;

Take moment about B;

MB=0VCB×24+(2×24×242)84+84=0VCB=57624VCB=24kips

Determine the shear VBC using the relation;

V=0VBC+VCB(2×24)=0VBC=4824VBC=24kips

Hence, the vertical reaction at support B is 41.3kips_.

Consider span CD;

Determine the shear VDC using the relation;

Take moment about C;

MC=0VDC×16+(30×8)84+48=0VDC=20416VDC=12.7kips

Hence, the vertical reaction at D is 12.7kips_.

Determine the shear VCD using the relation;

V=0VDC+VCD30=0VCD=3012.7VCD=17.3kips

Hence, the vertical reaction at C is 41.3kips_.

Shear force calculation;

VA=12.7kipsV8ft(left)=12.7kipsV8ft(right)=12.730V8ft(right)=17.3kipsVB(left)=17.3kipsVB(right)=12.7+41.330VB(right)=24kips

V24ft=12.7+41.330(2×12)V24ft=0VC(left)=12.7+41.330(2×24)VC(left)=24kipsVC(right)=12.7+41.3+41.330(2×24)VC(right)=17.3kips

V48ft(left)=12.7+41.3+41.330(2×24)V48ft(left)=17.3kipsV48ft(right)=12.7+41.3+41.330(2×24)30V48ft(right)=12.7kipsVD(left)=12.7+41.3+41.330(2×24)30VD(left)=12.7kipsVD(right)=12.7+41.3+41.330(2×24)30+12.7VD(right)=0

Bending moment calculations;

MA=48kips-ftM8ft=48+(12.7×8)M8ft54kips-ftMB=48+(12.7×16)(30×8)MB=84.8kips-ftM28ft=48+(12.7×28)(30×20)(2×12×122)+(41.3×12)M28ft=59.2kips-ft

MC=48+84+(12.7×40)(30×32)(2×24×242)+(41.3×24)MC=1,584+1,499.2MC=84.8kips-ftM48ft=48+84+(12.7×48)(30×40)(2×24×(242+8))+(41.3×32)+(41.3×8)M48ft=2,208+2,261.6M48ft54kips-ft

MD(left)=48kips-ftMD(right)=[(12.7×56)(30×48)(2×24×(242+16))+(41.3×40)+(41.3×16)(30×8)]MD(right)=3,024+3,024MD(right)=0

Consider a section at a distance of x from support A within 8 ft;

Mx=048+12.7x=0x=4812.7x=3.778ftfromsupportA

Similarly, the bending moment is zero from support D at 3.778 ft.

Consider span AB and DC;

Consider a section at a distance of x (within 16 ft) to determine the where the bending moment is equal to zero;

Mx=012.7x4830(x8)=012.7x4830x+240=0x=19217.3x=11.09ftfromsupportA

Similarly, the bending moment is zero from support D at 11.09 ft.

Consider span BC;

Consider a section at a distance of x (within 24 ft) to determine the where the bending moment is equal to zero;

Mx=024×x842×x×x2=0x2+24x84=0x=4.25ftfromsupportB

Similarly, the bending moment is zero from support C at 4.25 ft.

Show the shear force and bending moment diagram as in Figure (2).

Fundamentals of Structural Analysis, Chapter 10, Problem 11P , additional homework tip  2

The maximum deflection occurs between span B and C at point E.

Determine the deflection at E under the load using the relations;

ΔE=84(4.25)3E(1.5I)4.254+2(7.75)603E(1.5I)(1238×7.75)=1,795.08EI

Hence, the maximum deflection at E under the load is 1,795.08EI_.

Show the deflected shape of the frame as in Figure (3).

Fundamentals of Structural Analysis, Chapter 10, Problem 11P , additional homework tip  3

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