Problem 3.19. Multicell bridge

A multicell bridge cross section given in the Figure is subjected to a torque, T = 25 kNm. Determine the rate of twist of the cross section and the shear flow in the walls. Shear modulus is: G = 78 GPa. Data given in the Figure are the sizes of the midline.

Solve Problem

Solve

Rate of twist, ϑ1m=×10-5

Shear flow in the exterior cells, q1[kN/m]=q3[kN/m]= 

Shear flow in the exterior cells, q2 [kN/m]=

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Steps

Step by step

Step 1.  Cut cells, draw shear flows and summarize them to get torque resultant.

Check shear flows

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T=2A1q1+2A2q2+2A3q3

where

A1=A3=l1+l32l4=1.5×2=3 m2A2=l5l4=2×2=4 m2

Step 2.  Determine the relative displacement at the cuts from the sear flow.

Check relative displacement

The length of the inclined wall is l2=12+22=2.24 m.

Cell 1

qGtds=1Gq1l1t2+q1l2t1+q1l3t2+q1q2l4t1=49.86q1G20q2G

Cell 2

qGtds=1G2q2l5t2+q2q1l4t1+q2q3l4t1=20q1G+50q1G20q3G

Cell 3

qGtds=1Gq3l1t2+q3l2t1+q3l3t2+q3q2l4t1=q2G20+49.86q3G

Step 3.  The relative displacement at the cuts must be zero. Write the compatibility condition.

Check compatibility condition

For each cells (i=1,2,3):ϑ2Ai+qi1Gtds=0thus1G49.8620020502002049.86q1q2q3=ϑ686      q1q2q3

Step 4.  Determine shear flows from the above condition for unit rate of twist, ϑ=1

Check shear flows from unit rate of twist

G= 78×106 kNm2

q10q20q30=78×10649.8620020502002049.861686=21.229.421.2×106 kN

Step 4.  Calculate rate of twist and the shear flows from the given torque.

Check result

Ratio of the given torque and the resultant torque arises from unit rate of twist gives the rate of twist arising from the given torque.

T0=2A1q10+2A2q20+2A3q30=4.898×108 kNm2T=25kNm=ϑT0      ϑ=TT0=254.898×108 =5.10×1081m

q1q2q3=q10q20q30ϑ=21.229.421.2×106×5.10×108=1.0821.5021.082kNm

Results

Show worked out solution

First cells are cut, we draw shear flows and we summarize them to get torque resultant.

This image has an empty alt attribute; its file name is 3_19_2.jpg

T=2A1q1+2A2q2+2A3q3

where

A1=A3=l1+l32l4=1.5×2=3 m2A2=l5l4=2×2=4 m2

Then we determine the relative displacement at each cut from the shear flow. The length of the inclined wall is l2=12+22=2.24 m.

Cell 1

qGtds=1Gq1l1t2+q1l2t1+q1l3t2+q1q2l4t1=49.86q1G20q2G

Cell 2

qGtds=1G2q2l5t2+q2q1l4t1+q2q3l4t1=20q1G+50q1G20q3G

Cell 3

qGtds=1Gq3l1t2+q3l2t1+q3l3t2+q3q2l4t1=q2G20+49.86q3G

The relative displacement at the cuts must be zero. The compatibility condition is

For each cells (i=1,2,3):ϑ2Ai+qi1Gtds=0thus1G49.8620020502002049.86q1q2q3=ϑ686      q1q2q3

Shear flows can be determined from the above condition for unit rate of twist, ϑ=1 G=78×106 kNm2

q10q20q30=78×10649.8620020502002049.861686=21.229.421.2×106 kN

Ratio of the given torque and the resultant torque arises from unit rate of twist gives the rate of twist arising from the given torque.

T0=2A1q10+2A2q20+2A3q30=4.898×108 kNm2T=25kNm=ϑT0      ϑ=TT0=254.898×108 =5.10×1081m

q1q2q3=q10q20q30ϑ=21.229.421.2×106×5.1×108=1.0821.5021.082kNm