Concept explainers
Find the slope θB&θD and deflection ΔB&ΔD at point B and D of the given beam using the moment area method.
Answer to Problem 60P
The slope θB at point B (left) of the given beam using the moment area method is 0.0033 rad_.
The deflection ΔB at point B (left) of the given beam using the moment area method is 0.39 in.(↑)_.
The slope θB at point B (right) of the given beam using the moment area method is −0.00263 rad_.
The slope θD at point D of the given beam using the moment area method is 0.0071 rad_.
The deflection ΔD at point D of the given beam using the moment area method is 0.62 in.(↓)_.
Explanation of Solution
Given information:
The Young’s modulus (E) is 30,000 ksi.
The moment of inertia of the section AB is (I) is 4,000 in.4.
The moment of inertia of the section BD is (I) is 3,000 in.4.
Calculation:
Consider flexural rigidity EI of the beam is constant.
To draw a M/EI diagram, the reactions are calculated by considering all the loads acting in the beam. The calculation of positive moment at point A is calculated by simply considering the point load of 35 kips and the reactions calculated by considering all the loads acting in the beam.
Show the free body diagram of the given beam as in Figure (1).
Refer Figure (1),
Consider upward is positive and downward is negative.
Consider clockwise is negative and counterclowise is positive.
Refer Figure (1),
Consider reaction at A and C as RA and RC.
Take moment about point B.
Determine the reaction at D;
RC×(8)−(35×16)=0RC=5608RC=70 kips
Determine the reaction at support A;
∑V=0RA+RC−(2.5×16)−35=0RA=75−70RA=5 kips
Determine the moment at A:
MA=−(35×32)+(70×24)−(25×16×162)=−240 kips-ft=240 kips-ft(Clockwise)
Show the reaction of the given beam as in Figure (2).
Determine the bending moment at B;
MB=70×8−35×16=560−560=0
Determine the bending moment at C;
MC=−(35×8)=−280 kips-ft
Determine the bending moment at D;
MD=−(5×32)−(70×8)−240+(2.5×16×(162+16))=−720+240+960=0
Determine the positive bending moment at A using the relation;
MA=−(35×32)+70×24=−1,120+1,680=560 kips-ft
Show the reaction and point load of the beam as in Figure (3).
Determine the value of M/EI;
MEI=560 kips-ftEI
Substitute 4I3 for I.
MEI=560 kips-ftE(4I3)=1,680 kips-ft4EI=420 kips-ftEI
Show the M/EI diagram of the beam as in Figure (4).
Show the conjugate beam as in Figure (5).
Determine the support reaction at support B;
RB×8+1EI[13×240×16×(8+34×16)−12×16×420×(8+23×16)+12×8×280×(13×8)]=0RB=25,600−62,731+2,9878EIRB=−4,268 kips-ft2EIRB=4,268 kips-ft2EI(↓)
Determine the shear force at B (left) using the relation;
SB(left)=[12×b1×h1−13×b2×h2]
Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, and −240EI for h2.
SB(left)=[12×16×420EI−13×16×240EI]=2,080 kips-ft2EI
Determine the slope at B (left) using the relation;
SB(left)=2,080 kips-ft2EI
Substitute 30,000 ksi for E and 4,000 in.4 for I.
θB(left)=2,080 kips-ft2×(12 in.1 ft)230,000×3,000=0.0033 rad
Hence, the slope at B (left) is 0.0033 rad_.
Determine the slope at B (right) using the relation;
θB(right)=θB(left)−RB
Substitute 2,080 kips-ft2EI for θB(left) and 4,268 kips-ft2EI for RB.
θB(right)=2,080 kips-ft2EI−4,268 kips-ft2EI=−2,188 kips-ft2EI
Substitute 30,000 ksi for E and 4,000 in.4 for I.
θB(right)=−2,188 kips-ft2×(12 in.1 ft)230,000×4,000=−0.00263 rad
Hence, the deflection at B (right) is −0.00263 rad_.
Determine the bending moment at B using the relation;
M=[12×b1×h1×(23×b1)+13×b2×h2×(34×b2)]
Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, (−240EI) for h2.
MB=[12×16×420EI×(23×16)−13×16×(240EI)×(34×16)]=35,840−15,360EI=20,480 kips-ft2EI
Determine the deflection at B using the relation;
MB=20,480 kips-ft2EI
Substitute 30,000 ksi for E and 4,000 in.4 for I.
MB=20,480 kips-ft2×(12 in.1 ft)330,000×3,000=0.39 in.(↑)
Hence, the deflection at B is 0.39 in.(↑)_.
Determine the shear force at D using the relation;
SD=[12×b1×h1+13×b2×h2+RB+12×b3×h3]
Here, b is the width and h is the height of respective triangle and parabola.
Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, −240EI for h2, 4,268 kips-ft2EI for RB, 16 ft for b3, and −280EI for h3.
SD=[12×16×420EI−13×16×(240EI)−4,268−12×16×(280EI)]=3,360−1,280−4,268−2,240EI=−4,427 kips-ft2EI
Determine the slope at D using the relation;
θD=−4,427 kips-ft2EI
Substitute 30,000 ksi for E and 3,000 in.4 for I.
θD=−4,427 kips-ft2×(12 in.1 ft)230,000×3,000=−0.0071 rad
Hence, the deflection at D is −0.0071 rad_
Determine the bending moment at D using the relation;
MD=[12×b1×h1×(16+23×b1)+13×b2×h2×(16+34×b2)+(RB×16)+12×b3×h3×(24×b3)]
Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, −240EI for h2, 4,268 kips-ft2EI for RB, 16 ft for b3, and −280EI for h3.
MD=[12×16×(420EI)(16+23×16)−13×16×(240EI)(16+34×16)−(4,268×16)−12×16×(280EI)×(24×16)]=89,611−35,840−68,288−17,920EI=−32,437 kips-ft3EI
Determine the deflection at D using the relation;
ΔD=−32,437 kips-ft3EI
Substitute 30,000 ksi for E and 3,000 in.4 for I.
ΔD=−32,437 kips-ft3×(12 in.1 ft)330,000×3,000=−0.62 in.=0.62 in.(↓)
Hence, the deflection at D is 0.62 in.(↓)_
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