Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P16.1–P16.5 by using the moment-distribution method. FIG. P16.1

Find the reaction and plot the shear and bending moment diagram.

Fixed end moment:

Formula to calculate the relative stiffness for fixed support IL and for roller support (34)(IL).

Formula to calculate the fixed moment for point load with equal length are PL8.

Formula to calculate the fixed moment for point load with equal length are PL8.

Formula to calculate the fixed moment for point load with unequal length are Pab2L2 and Pa2bL2.

Formula to calculate the fixed moment for UDL is WL212.

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the free body diagram of the entire beam as in Figure 1.

Refer Figure 1,

Calculate the relative stiffness KCA for part CA of the beam:

KCA=I20 ft+10 ft=I30 

Calculate the relative stiffness KCE for part CE of the beam:

KCE=I15 ft+15 ft=I30

In the above beam, only joint C is free to rotate. Hence, calculate the distribution factor at joint C.

Calculate the distribution factor DFCA for member AC of the beam.

DFCA=KCAKCA+KCE

Substitute I30  for KCA and I30  for KCE.

DFCA=I30 I30 +I30 =0.5

Calculate the distribution factor DFCA for part CE of the beam.

DFCE=KCEKCA+KCE

Substitute I30  for KCA and I30  for KCE.

DFCE=I30 I30 +I30 =0.5

Check for sum of distribution factor:

DFCE+DFCA=1

Substitute 0.5 for DFCE and 0.5 for DFCA.

0.5+0.5=1

Hence, OK.

Calculate the fixed end moment for AC.

FEMAC=18×20×102(30)2=40 k⋅ft

Calculate the fixed end moment for CA.

FEMCA=−18×202×10(30)2=−80 k⋅ft

Calculate the fixed end moment for CE.

FEMCE=10×308=37.5k⋅ft

Calculate the fixed end moment for EC.

FEMEC=−10×308=−37.5k⋅ft

Show the calculation of final moments using moment distribution method as in Table 1.

Consider the member AC of the beam:

Show the free body diagram of the member AC as in Figure 2.

Calculate the vertical reaction at the left end of the joint C by taking moment about point A.

+↺∑MA=0Cy,L(30)−18×(20)+50.625−58.75=0Cy,L(30)=368.125Cy,L=368.12530Cy,L=12.271 k

Calculate the horizontal reaction at point A by resolving the horizontal equilibrium.

+→∑Fx=0Ax=0

Calculate the vertical reaction at point A by resolving the vertical equilibrium.

+↑∑Fy=0Ay−18+Cy,L=0Ay−18+12.271=0Ay=5.729 k↑

Consider the member CE of the beam:

Show the free body diagram of the member CE as in Figure 3.

Calculate the vertical reaction at the right end of the joint C by taking moment about point E.

+↺∑ME=0Cy,R(30)+10×(15)−26.875+58.75=0Cy,R(30)=181.875Cy,R=181.87530Cy,R=6.063 k↑

Calculate the horizontal reaction at point E by resolving the horizontal equilibrium.

+→∑Fx=0Ex=0

Calculate the vertical reaction at point E by resolving the vertical equilibrium.

+↑∑Fy=0Ey−10+Cy,R=0Ey−10+6.063=0Ey=3.937 k↑

Calculate the total reaction at point C.

Cy=Cy,R+Cy,L

Substitute 6.063 k for Cy,R and 12.271 k for Cy,L.

Cy=6.063+12.271=18.334 k↑

Show the reaction of the beam in Figure 4.

Refer Figure 4,

Shear diagram:

Point A:

SA,L=0SA,R=5.729 k

Point B:

SB=5.729−18=−12.271 k

Point C:

SC,L=−12.271 kSC,R=−12.271+18.334=6.063 k

Point D:

SD,L=6.063kSD,R=6.063−10=−3.937 k

Point E:

SE,L=−3.937 kSE,R=−3.937 k+3.937=0 k

Plot the shear force diagram of the beam as in Figure 5.

Refer Figure 4,

Bending moment diagram:

Point A:

MA=−50.625 k⋅ft

Point B:

MB=−50.625+(5.729×20)=63.955 k⋅ft

Point C:

MC=−58.75 k⋅ft≃58.8 k⋅ft

Point D:

MD=−58.8+(6.063×15)=32.145 k

Point E:

ME=−26.875 k⋅ft≃26.9 k⋅ft

Plot the bending moment diagram of the beam as in Figure 6.