EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 9780100257061
Author: BEER
Publisher: YUZU
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Textbook Question
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Chapter 9.2, Problem 11P

For the beam and loading shown, (a) express the magnitude and location of the maximum deflection in terms of w0, L, E, and I. (b) Calculate the value of the maximum deflection, assuming that beam AB is a W18 × 50 rolled shape and that w0= 4.5 kips/ft, L = 18 ft, and E = 29 ×106 psi.

Chapter 9.2, Problem 11P, For the beam and loading shown, (a) express the magnitude and location of the maximum deflection in

Fig. P9.11

(a)

Expert Solution
Check Mark
To determine

The magnitude and location of the maximum deflection in terms of w0, L, E, and I.

Answer to Problem 11P

The location of the maximum deflection xm is 0.481L_.

The magnitude and location of the maximum deflection in terms of w0, L, E, and I is 0.00652w0L4EI_.

Explanation of Solution

Given that:

The length (L) of the beam is 18ft.

The load w0 is 4.5kips/ft.

The young’s modulus E is 29×106psi.

Calculation:

Sketch the free body diagram of beam as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 9.2, Problem 11P , additional homework tip  1

Find the reactions of the beam.

RA+RB=12×wo×L

Take the moment at B.

RA×L(12×wo×L)(23×L)=0RA×LwoL23=0RA=woL23×1L=woL3

Find the reaction at B.

RA+RB=12×wo×LwoL3+RB=12×wo×LRB=woL2woL3=woL6

Take the section 1-1 at x distance from A as shown in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 9.2, Problem 11P , additional homework tip  2

Consider a section xx at a distance x from a.

Sketch the section x-x as shown in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 9.2, Problem 11P , additional homework tip  3

Calculate the intensity of loading w at the section x using similar triangle method as shown below:

wLx=w0Lw=w0L(Lx)

Find the shear force using the expression as follows:

dVdx=w=wo(Lx)L

Find the shear force using integration:

dVdx=wo(Lx)LV=woL(Lxx22)+CV=dMdx

Find the moment using the relation as follows:

dMdx=woL(Lxx22)+CVM=woL(Lx22x36)+CVx+CM (1)

Apply the boundary conditions:

When x=0, the moment M=0, substitute in Equation (1):

M=woL(Lx22x36)+CVx+CM0=woL(00)+0+CMCM=0

When x=L, the moment M=0.

Substitute 0 for CM and boundary condition in Equation (1):

M=woL(Lx22x36)+CVx+CM0=woL(L(L)22L36)+CVL+0=woL(L32L36)+CVL=woL(2L36)+CVL

CVL=woL(L33)CV=woL(L33)(1L)=woL3

Write the moment Equation:

EId2ydx2=M=woL(Lx22x36)+CVx+CM

Substitute woL3 for CV and 0 for CM.

EId2ydx2=woL(Lx22x36)+(w0L3)x+0=woL(Lx22x36)+(w0L3)x×(LL)=woL(Lx22x36L2x3)=woL(L2x3Lx22+x36) (2)

Integrate the Equation (2).

EId2ydx2=woL(L2x3Lx22+x36)EIdydx=woL(L2x26Lx36+x424)+C1 (3)

Integrate the Equation (3).

EIdydx=woL(L2x26Lx36+x424)+C1EIy=woL(L2x318Lx424+x5120)+C1x+C2 (4)

EIy=woL(L2x318Lx424+x5120)+C1x+C2

Apply the boundary condition in C2 as follows:

At x=0, y=0

Find the C2 value as shown follows:

Substitute 0 for x and 0 for y in Equation (4).

EIy=woL(L2(0)318L(0)424+(0)5120)+C1(0)+C2=woL(00+0)+0+C2C2=0

Apply the boundary condition in C1 as follows:

At x=L, y=0

Find the C1 value as shown follows:

Substitute 0 for x and 0 for y in Equation (4).

EI(0)=woL(L2(L)318L(L)424+(L)5120)+C1(L)+C20=woL(L518L524+(L)5120)+C1(L)+0C1(L)=woL(L545)C1(L)=woL445

C1=woL345

Substitute woL345 for C1 and 0 for C2 in Equation (4).

EIy=woL(L2x318Lx424+x5120)woL345x+0y=woEIL(118L2x3124Lx4+1120x5145L4x) (5)

Differentiate with respect to x in Equation (5).

dydx=woEIL(16L2x216Lx3+124x4145L4) (6)

To find the location of maximum deflection:

dydx=0

f=16L2xm216Lxm3+124xm4145L4

Consider the function z=xmL.

f(z)=16z216z3+124z4145 (7)

Differentiate with respect to z in Equation (7).

f(z)=16z216z3+124z4145dfdz=13zz22+16z3

Find the value z using Newton-Raphson method as follows:

z=z0f(z0)dfdz (8)

Show the calculated values of f(z), dfdz, and z values as shown in Table 1.

z0f(z)dfdzz
0.22-0.015830.0509080.53100
0.24-0.014790.0535040.51639
0.26-0.013690.0557960.50544
0.28-0.012560.0577920.49730
0.3-0.011380.05950.49134
0.32-0.010180.0609280.48708
0.34-0.008950.0620840.48415
0.36-0.00770.0629760.48224
0.38-0.006430.0636120.48111
0.4-0.005160.0640.48056
0.42-0.003870.0641480.48039
0.44-0.002590.0640640.48045
0.46-0.001310.0637560.48059
0.48-4.2E-050.0632320.4807
0.50000.00120.06250.4806
0.520.0024560.0615680.48010

Refer to table: 1.

The value of xm is 0.481L.

Find the value of ym using the relation as follows:

Substitute 0.481L for xm in Equation (5).

ym=woEIL(118L2(0.481L)3124L(0.481L)4+1120(0.481L)5145L4(0.481L))=woL4EIL(118(0.4807)3124(0.4807)4+1120(0.4807)5145(0.4807))=woL4EIL(0.006170.00222+0.000213880.01068)=0.00652woL4EIL

Therefore, he magnitude of the maximum deflection in terms of w0, L, E, and I is 0.00652w0L4EI_.

Therefore, the location of maximum deflection is 0.481L_.

(b)

Expert Solution
Check Mark
To determine

The value of maximum deflection.

Answer to Problem 11P

The value of maximum deflection is 0.229in._

Explanation of Solution

Calculation:

Convert kips/ft to lb/in.

w0=4.5kips/ft=450012=375lb/in.

The rolled shape section W18×50,

The value of I is 800in4.

Find the maximum deflection using the relation:

ym=0.00652w0L4EIL (9).

Substitute 375lb/in. for w0, 18ft for L, and 800in4 for I in Equation (9).

ym=0.00652375×18ft(12in1ft)29×106(800)=0.00652375×(216)429×106(800)=0.229in.

Thus, the value of maximum deflection is 0.229in._

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