Find the equations for slope and deflection of the beam using direct integration method.

Question

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Answer 1

Question

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

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Answer 2

Question

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

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Answer 3

Question

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

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Answer 4

Question

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

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Answer 5

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

Answer 6

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

Answer 7

Chapter 6, Problem 3P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Answer 8

Expert Solution & Answer

Answer to Problem 3P

The equation for slope for segment AB is $wx2EI(a2−L2+(L−a)x)_$ .

The equation for deflection for segment AB is $wx22EI(a2−L22+(L−a)x3)_$ .

The equation for slope for segment BC is $w2EI(xL(x−L)−x33+a33)_$ .

The equation for deflection for segment BC is $w2EI(x2L(x3−L2)−x412−a412+a3x3)_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Draw the free body diagram of the beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 1

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$ΣFy=0Ay−w(L−a)=0Ay=w(L−a)$

Find the moment at A.

Consider anticlockwise moment as positive.

$0=MA−w×(L−a)×[(L−a)2+a]MA=w(L−a)[(L−a+2a2)]MA=w(L−a)(L+a2)MA=w2(L2−a2)$

Segment AB: $0≤x≤a$ .

Consider a section $X1X1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X1X1$ consider in the segment AB as shown in Figure 2.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 2

Refer Figure (2),

Take the moment at section $X1X1$ .

$M=−w2(L2−a2)+w×(L−a)(x)$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w(L−a)x]$ (1)

Find the equation for slope $(θ)$ .

Integrate Equation (1) with respect to x.

$dydx=θ=∫MEIdxθ=∫−w2(L2−a2)+w×(L−a)(x)EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)+C1$ (2)

Find the equation for deflection $(θ)$ .

Integrate again Equation (2) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+C1x+C2$ (3)

Find the integration constants $C1 and C2$ :

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$ .

$0=M×022EI+C1×0+C2C2=0$

Apply boundary conditions in Equation (1):

At $x=0$ and $θ=0$ .

$0=−w2(L2−a2)×0+w×(L−a)(022)+C1C1=0$

Find the equation for slope of segment AB.

Substitute 0 for $C1$ in Equation (2).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)+0EIθ=−wx2(L2−a2)+wx22(L−a)θ=wx2EI[a2−L2+(L−a)x]$

Thus, the equation for slope of segment AB is $wx2EI[a2−L2+(L−a)x]_$ .

Find the equation for deflection of segment AB.

Substitute 0 for $C1$ and 0 for $C2$ in Equation (3).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)+0×x+0EIy=−w2(L2−a2)x22+w×(L−a)(x36)EIy=−wx22(L2−a22)+wx36(L−a)y=wx22EI(a2−L22+(L−a)x3)$

Thus, the equation for deflection is $wx22EI(a2−L22+(L−a)x3)_$ .

Segment BC: $a≤x≤L$ .

Consider a section $X2X2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X2X2$ consider segment BC as shown in Figure 3.

Structural Analysis, SI Edition, Chapter 6, Problem 3P , additional homework tip 3

Refer Figure 3.

Write the equation for bending moment at section $X2X2$ .

$M=−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2$

Write the equation for $MEI$ .

$d2ydx2=MEI=1EI[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]$ (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$dydx=θ=∫MEIdxθ=∫[−w2(L2−a2)+w×(L−a)(x)−w2(x−a)2]EIdxEIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+C3$ (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+C3x+C4$ (6)

Find the integration constants $C3 and C4$ :

Apply boundary conditions in Equation (3):

At $x=0$ , $(L−a)=0$ , and $θB,Left=θB,Right$ .

$EIθ=−w2(a33+a2×a−a×a2)+C30=−w2(a33+a3−a3)+C30=−wa36+C3C3=wa36$

Apply boundary conditions in Equation (4):

At $x=0$ , $(L−a)=0$ , and $yB,Left=yB,Right$ .

$0=−w2(a412+a2a22−a×a33)+wa36×a+C40=−w2(a4+6a4−4a412)+wa46+C40=−w2(3a412)+wa46+C40=−wa48+wa46+C4$

$0=−6wa4+8wa448+C40=wa424+C4C4=−wa424$

Find the equation for slope of segment BC.

Substitute $wa36$ for $C3$ in Equation (5).

$EIθ=−w2(L2−a2)x+w×(L−a)(x22)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x−a2x)+w2(Lx2−ax2)−w2(x33+a2x−ax2)+wa36EIθ=−w2(L2x)+w2(Lx2)−w2(x33)+wa36EIθ=w2[L2x+Lx2−x33+a33]$

$θ=w2EI[xL(x−L)−x33+a33]$

Thus, the equation for slope of segment BC is $w2EI[xL(x−L)−x33+a33]_$ .

Find the equation for deflection of segment BC.

Substitute $wa36$ for $C3$ and $−wa424$ for $C4$ in Equation (6).

$EIy=−w2(L2−a2)x22+w×(L−a)(x36)−w2(x412+a2x22−ax33)+wa36x−wa424EIy=−w2(L2x22−a2x22)+w(Lx36−ax36)−w2(x412+a2x22−ax33)+wxa36−wa424EIy={w2(−L2x22)+w2(a2x22)+w2(Lx33−ax33)+w2(−x412+ax33)+w2(−a2x22)+wxa36−wa424}y=w2(−L2x22+Lx33−ax33−x412+ax33+xa33−a412)$ $y=w2[x2L(x3−L2)−x412−a412+xa33]=w2EI[x2L(x3−L2)−x412−a412+xa33]$

Thus, the equation for deflection of segment BC is $w2EI[x2L(x3−L2)−x412−a412+xa33]_$ .

Answer 12

Ch. 6 - Prob. 1P Ch. 6 - Prob. 2P Ch. 6 - Prob. 3P Ch. 6 - Prob. 4P Ch. 6 - Prob. 5P Ch. 6 - Prob. 6P Ch. 6 - Prob. 7P Ch. 6 - Prob. 8P Ch. 6 - Prob. 9P Ch. 6 - Prob. 10P

Find the equations for slope and deflection of the beam using direct integration method.

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Find the equations for slope and deflection of the beam using direct integration method.

Answer to Problem 3P

Explanation of Solution

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