Given information:
The Young’s modulus (E) is 29,000 ksi.
The moment of inertia (I) is 6,000 in.4.
Calculation:
Consider flexural rigidity EI of the beam is constant.
Show the free body diagram of the given beam as in Figure (1).
Refer Figure (1),
Consider upward is force is positive and downward force is negative.
Consider clockwise moment is negative and counterclockwise moment is positive.
Split the given beam into two sections such as AC and CE.
Consider the portion CE:
Draw the free body diagram of the portion CE as in Figure (2).
Refer Figure (2),
Consider a reaction at C and take moment about point C.
Determine the reaction at E;
RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips
Determine the reaction at support A;
∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips
Show the reaction of the given beam as in Figure (3).
Refer Figure (3),
Determine the moment at A:
MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft
Determine the bending moment at B;
MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft
Determine the bending moment at C;
MC=(9.58×24)+250−(40×12)=479.92−480≃0
Determine the bending moment at D;
MD=(9.58×12)+250≃365 kips-ft
Determine the bending moment at E;
ME=250 kips-ft
Show the M/EI diagram for the given beam as in Figure (4).
Show the elastic curve diagram as in Figure (5).
Refer Figure (4),
Determine the slope at B;
θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)
Here, b is the width and h is the height of respective triangle.
Substitute 12 ft for b1, 1,630EI for h1, 12 ft for b2, and 365EI for h2.
θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI
Substitute 29,000 ksi for E and 6,000 in.4 for I.
θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)
Hence, the slope at B is 0.0099 rad(Clockwise)_.
Determine the deflection between A and B using the relation;
ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]
Here, b1 is the width of rectangle, h1 is the height of the rectangle, b2 is the width of the triangle, and h2 is the height of the triangle.
Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, and (1,630EI−365EI) for h2.
ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI
Substitute 29,000 ksi for E and 6,000 in.4 for I.
ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)
Hence, the deflection at B is 0.86 in.(↓)_.
To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.
Determine the deflection at C and A using the relation;
ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]
Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, (1,630EI−365EI) for h2, 12 for b3, and 365EI for h3.
ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI
Determine the deflection between C and E using the relation;
ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]
Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, 250EI for h2, 12 ft for b3, and (365EI−250EI) for h3.
ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI
Determine the slope at E using the relation;
θE=ΔC+ΔCELCE
Here, LCE is the length between C and E.
Substitute −248,160 kips-ft3EI for ΔC, −82,560 kips-ft3EI for ΔCE, and 24 ft for LCE.
θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI
Determine the slope between D and E using the relation;
θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)
Here, b is the width and h is the height of respective triangle.
Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, and 250EI for h2.
θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI
Determine the slope at D using the relation;
θD=θE−θDE
Substitute −3,690 kips-ft2EI for θE and −3,690 kips-ft2EI for θDE.
θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI
Substitute 29,000 ksi for E and 6,000 in.4 for I.
θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)
Hence, the slope at D is 0.0084 rad(Counterclockwise)_.
Determine the deflection between D and E using the relation;
ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]
Here, b is the width and h is the height of respective rectangle and triangle.
Substitute 12 ft for b1, 250EI for h1, 12 for b2, and (365EI−250EI) for h2.
ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI
Determine the deflection at D using the relation;
ΔD=LDEθE−ΔDE
Substitute 12 ft for LDE, −13,780 kips-ft2EI for θE, and −20,760 kips-ft3EI for ΔDE.
ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI
Substitute 29,000 ksi for E and 6,000 in.4 for I.
ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)
Hence, the deflection at point D is 1.44 in.(↓)_.