6.33 and 6.34 Use the moment-area method to determine the slopes and deflections at point B and D of the beam shown FIG.P6.34, P6.60

Find the slope θB&θD and deflection ΔB&ΔD at point B and D of the given beam using the moment area method.

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 Young’s modulus (E) 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 C;

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 reactions 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;

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 elastic curve as in Figure (5).

The slope at point B can be calculated by evaluating the change in slope between A and B.

Express the change in slope using the first moment-area theorem as follows:

θB(left)=θAB=Area of the M/EI between A and B=Area of triangle+Area of parabola=[12×b1×h1−13×b2×h2]

Here, b is the width and h is the height of the respective triangle and rectangle.

Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, and −240EI for h2.

θB(left)=[12×16×420EI−13×16×240EI]=2,080 kips-ft2EI

Determine the slope B (left) using the relation;

θB(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 deflection at B using the relation;

ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=[MEI(Area of triangle)+MEI(Area of parabola)]

Substitute 420EI for MEI, (12×16×(23×16)) for area of triangle, (−240EI) for MEI, and (13×16×(34×16)) for area of triangle.

ΔBA=[420EI(12×16×(23×16))+(−240EI)(13×16×(34×16))]=1EI(35,840−15,360)=20,480 kips-ft3EI

Determine the deflection at B (left) using the relation;

ΔBA=20,480 kips-ft3EI

Substitute 30,000 ksi for E and 3,000 in.4 for I.

ΔBA=20,480 kips-ft3×(12 in.1 ft)330,0000×3,000=0.39 in.(↑)

Hence, the deflection at B (left) is 0.39 in.(↑)_.

Determine the deflection between B and C using the relation;

ΔBC=Moment of the area of the M/EI diagram between B and C about C=[MEI(Area of triangle)×(23×b)]=[12×b×h×(23×b)]

Here, b is the width and h is the height of the triangle