08-parcial 1 (shaums)

7/31/2019 08-Parcial 1 (Shaums)

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LEDEZMA FLORES SAMUEL

FUNDAMENTOS DE MECANICA DE SOLIDOS Pgina 14

PARCIAL 1 SCHAUMS + AULA POLI

SERIE SCHAUMS CAPITULO 6

Solucin: A) EQUILIBRIUM EQUATIONS

Sum of Vertical Forces(Reactions Ay and By assumed positive upward)

Ay+ By= 120.00 lb/ft 10.00 ft

Sum of Moments about A (Right hand rule sign convention)(10.00 ft) By= 120.00 lb/ft 10.00 ft 5.00 ft

The total magnitude of the distributed load in this region is 120.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 10.00 ft is -1,200.00 lb (i.e., -120.00 lb/ft 10.00 ft ). The change inshear between x = 0.00 ft and x = 10.00 ft is equal to the area under the load diagram between these two points. Atx = 0.00 ft, the shear force is 600.00 lb. Adding -1,200.00 lb to this value gives a shear force of V = -600.00 lb at x= 10.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -120.00 lb/ft).Since its slope is constant, the shear curve is linear in this region.


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Solution: A) EQUILIBRIUM EQUATIONS


Ay+ By= 0.5(600.00 lb/ft) 12.00 ft

Sum of Moments about A (Right hand rule sign convention)(12.00 ft) By= 0.5(600.00 lb/ft)(12.00 ft)(8.00 ft)

The area under the load diagram between x = 0.00 ft and x = 12.00 ft is -3,600.00 lb. The change in shear betweenx = 0.00 ft and x = 12.00 ft is equal to the area under the load diagram between these two points. At x = 0.00 ft,the shear force is 1,200.00 lb. Adding -3,600.00 lb to this value gives a shear force of V = -2,400.00 lb at x = 12.00ft. The shear curve is parabolic in this region.


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Sum of Vertical Forces(Reaction Ay assumed positive upward)

Ay= + 600.00 N/m 2.00 m - 0.00 N

Sum of Moments about A (Right hand rule sign convention)(Reaction MA assumed positive clockwise)

MA= - 0.00 N-m - 600.00 N/m 2.00 m 3.00 m - 4,800.00 N-m - 0.00 N 0.00 m

The total magnitude of the distributed load in this region is 600.00 N/m, acting DOWNWARD. The area

under the load diagram between x = 2.00 m and x = 4.00 m is -1,200.00 N (i.e., -600.00 N/m 2.00 m ). Thechange in shear between x = 2.00 m and x = 4.00 m is equal to the area under the load diagram between these two

points. At x = 2.00 m, the shear force is 1,200.00 N. Adding -1,200.00 N to this value gives a shear force of V =

152.40E-06 N at x = 4.00 m. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w

= -600.00 N/m). Since its slope is constant, the shear curve is linear in this region.


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Ay+ By= - 0.00 N + 300.00 N/m 6.00 m - 0.00 N

Sum of Moments about A (Right hand rule sign convention)(6.00 m) By= - 0.00 N 0.00 m + 300.00 N/m 6.00 m 6.00 m - 2,700.00 N-m - 0.00 N 6.00 m

The total magnitude of the distributed load in this region is 300.00 N/m, acting DOWNWARD. The area

under the load diagram between x = 6.00 m and x = 9.00 m is -900.00 N (i.e., -300.00 N/m 3.00 m ). The change

in shear between x = 6.00 m and x = 9.00 m is equal to the area under the load diagram between these two points.

At x = 6.00 m, the shear force is 900.00 N. Adding -900.00 N to this value gives a shear force of V = 76.20E-06 N at

x = 9.00 m. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -300.00

N/m). Since its slope is constant, the shear curve is linear in this region.


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Ay+ By= 4,000.00 lb - 0.00 lb - 0.00 lb

Sum of Moments about A (Right hand rule sign convention)(4.00 ft) By= 4,000.00 lb 1.00 ft - 0.00 lb 0.00 ft - 0.00 lb 4.00 ft

At x = 1.00 ft, a concentrated load of4,000.00 lb acts downward on the beam. Concentrated forces create

discontinuities in the shear diagram. The load of4,000.00 lb causes the shear diagram to jump DOWN by 4,000.00

lb at x = 1.00 ft. Just slightly to the left ofx = 1.00 ft, the shear force is 3,000.00 lb. Adding -4,000.00 lb to this

value gives a shear force of-1,000.00 lb just slightly to the right ofx = 1.00 ft.


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Solucion: A) EQUILIBRIUM EQUATIONS

Sum of Vertical Forces(Reaction By assumed positive upward)

By= 0

Sum of Moments about B (Right hand rule sign convention)(Reaction MB assumed positive counterclockwise)

MB= - 200.00 lb-ft

A concentrated moment of200.00 lb-ft acts counterclockwise on the beam at x = 6.00 ft. While concentratedmoments don't directly affect the shear diagram, they cause discontinuities in the moment curve. Thecounterclockwise concentrated moment of200.00 lb-ft causes the moment diagram to jump DOWN by -200.00 lb-ftat x = 6.00 ft. Just slightly to the left ofx = 6.00 ft, the bending moment is 0.00 lb-ft. Adding -200.00 lb-ft to thisvalue gives a bending moment of-200.00 lb-ft just slightly to the right ofx = 6.00 ft.


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Ay+ By= 1.00 kN - 1.00 kN

Sum of Moments about A (Right hand rule sign convention)(3.00 m) By= - 1.00 kN 0.00 m - 1.00 kN 3.00 m + 3.00 kN-m

At x = 0.00 m, a concentrated load of1.00 kN acts downward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of1.00 kN causes the shear diagram to jump DOWN by 1.00 kN at x =0.00 m. Just slightly to the left ofx = 0.00 m, the shear force is 0.00 kN. Adding -1.00 kN to this value gives a shearforce of-1.00 kN just slightly to the right ofx = 0.00 m.

A concentrated moment of3.00 kN-m acts clockwise on the beam at x = 2.00 m. While concentrated momentsdon't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise concentratedmoment of3.00 kN-m causes the moment diagram to jump UP by -3.00 kN-m at x = 2.00 m. Just slightly to the leftofx = 2.00 m, the bending moment is -2.00 kN-m. Adding +3.00 kN-m to this value gives a bending moment of1.0000 kN-m just slightly to the right ofx = 2.00 m.

At x = 3.00 m, a concentrated load of1.00 kN acts upward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of1.00 kN causes the shear diagram to jump UP by 1.00 kN at x =3.00 m. Just slightly to the left ofx = 3.00 m, the shear force is -1.00 kN. Adding +1.00 kN to this value gives ashear force of0.00 kN just slightly to the right ofx = 3.00 m.


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Ay+ By= 300.00 lb/ft 1.00 ft + 300.00 lb/ft 1.00 ft

Sum of Moments about A (Right hand rule sign convention)(3.00 ft) By= - 300.00 lb/ft 1.00 ft 0.50 ft + 300.00 lb/ft 1.00 ft 3.50 ft - 150.00 lb-ft

The total magnitude of the distributed load in this region is 300.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 1.00 ft is -300.00 lb (i.e., -300.00 lb/ft 1.00 ft ). The change in shearbetween x = 0.00 ft and x = 1.00 ft is equal to the area under the load diagram between these two points. At x =0.00 ft, the shear force is -250.00E-06 lb. Adding -300.00 lb to this value gives a shear force of V = -300.00 lb at x =1.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -300.00 lb/ft).Since its slope is constant, the shear curve is linear in this region.

The total magnitude of the distributed load in this region is 300.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 4.00 ft and x = 5.00 ft is -300.00 lb (i.e., -300.00 lb/ft 1.00 ft ). The change in shearbetween x = 4.00 ft and x = 5.00 ft is equal to the area under the load diagram between these two points. At x =

4.00 ft, the shear force is 300.00 lb. Adding -300.00 lb to this value gives a shear force of V = 250.00E-06 lb at x =5.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -300.00 lb/ft).Since its slope is constant, the shear curve is linear in this region.

A concentrated moment of150.00 lb-ft acts counterclockwise on the beam at x = 2.50 ft. While concentrated

moments don't directly affect the shear diagram, they cause discontinuities in the moment curve. The

counterclockwise concentrated moment of150.00 lb-ft causes the moment diagram to jump DOWN by -150.00 lb-ft

at x = 2.50 ft. Just slightly to the left ofx = 2.50 ft, the bending moment is -75.00 lb-ft. Adding -150.00 lb-ft to this

value gives a bending moment of-225.00 lb-ft just slightly to the right ofx = 2.50 ft.


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Ay+ By= 10.00 kN/m 2.00 m + 12.00 kN

Sum of Moments about A (Right hand rule sign convention)(4.50 m) By= 10.00 kN/m 2.00 m 3.50 m + 12.00 kN 3.50 m

The total magnitude of the distributed load in this region is 10.00 kN/m, acting DOWNWARD. The area under theload diagram between x = 2.50 m and x = 3.50 m is -10.00 kN (i.e., -10.00 kN/m 1.00 m ). The change in shearbetween x = 2.50 m and x = 3.50 m is equal to the area under the load diagram between these two points. At x =2.50 m, the shear force is 7.11 kN. Adding -10.00 kN to this value gives a shear force of V = -2.89 kN at x = 3.50 m.

The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -10.00 kN/m). Since itsslope is constant, the shear curve is linear in this region.

At x = 3.50 m, a concentrated load of12.00 kN acts downward on the beam. Concentrated forces create

discontinuities in the shear diagram. The load of12.00 kN causes the shear diagram to jump DOWN by 12.00 kN at

x = 3.50 m. Just slightly to the left ofx = 3.50 m, the shear force is -2.89 kN. Adding -12.00 kN to this value gives

a shear force of-14.89 kN just slightly to the right ofx = 3.50 m.


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Ay+ By= 800.00 lb/ft 12.00 ft

Sum of Moments about A (Right hand rule sign convention)(17.00 ft) By= 800.00 lb/ft 12.00 ft 6.00 ft + 10,000.00 lb-ft

The total magnitude of the distributed load in this region is 800.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 12.00 ft is -9,600.00 lb (i.e., -800.00 lb/ft 12.00 ft ). The change inshear between x = 0.00 ft and x = 12.00 ft is equal to the area under the load diagram between these two points. Atx = 0.00 ft, the shear force is 5,623.53 lb. Adding -9,600.00 lb to this value gives a shear force of V = -3,976.47 lbat x = 12.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -800.00lb/ft). Since its slope is constant, the shear curve is linear in this region.

A concentrated moment of10,000.00 lb-ft acts clockwise on the beam at x = 14.00 ft. While concentrated

moments don't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise

concentrated moment of10,000.00 lb-ft causes the moment diagram to jump UP by -10,000.00 lb-ft at x = 14.00 ft.

Just slightly to the left ofx = 14.00 ft, the bending moment is 1,929.42 lb-ft. Adding +10,000.00 lb-ft to this value

gives a bending moment of11,929.41 lb-ft just slightly to the right ofx = 14.00 ft.


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Ay+ By= 2.00 kN/m 2.00 m - 2.00 kN - 2.00 kN

Sum of Moments about A (Right hand rule sign convention)(4.00 m) By= 2.00 kN/m 2.00 m 2.00 m - 2.00 kN 0.00 m - 2.00 kN 4.00 m

The total magnitude of the distributed load in this region is 2.00 kN/m, acting DOWNWARD. The area under theload diagram between x = 1.00 m and x = 3.00 m is -4.00 kN (i.e., -2.00 kN/m 2.00 m ). The change in shearbetween x = 1.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x =

1.00 m, the shear force is 2.00 kN. Adding -4.00 kN to this value gives a shear force of V = -2.00 kN at x = 3.00 m.The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -2.00 kN/m). Since itsslope is constant, the shear curve is linear in this region.

At x = 0.00 m, a concentrated load of2.00 kN acts upward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of2.00 kN causes the shear diagram to jump UP by 2.00 kN at x =0.00 m. Just slightly to the left ofx = 0.00 m, the shear force is 0.00 kN. Adding +2.00 kN to this value gives a

shear force of2.00 kN just slightly to the right ofx = 0.00 m.


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At x = 4.00 m, a concentrated load of2.00 kN acts upward on the beam. Concentrated forces create

discontinuities in the shear diagram. The load of2.00 kN causes the shear diagram to jump UP by 2.00 kN at x =

4.00 m. Just slightly to the left ofx = 4.00 m, the shear force is -2.00 kN. Adding +2.00 kN to this value gives a

shear force of0.0 kN just slightly to the right ofx = 4.00 m.


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Sum of Vertical Forces(Reaction By assumed positive upward)

By= 1.00 kN + 2.00 kN

Sum of Moments about B (Right hand rule sign convention)(Reaction MB assumed positive counterclockwise)

MB= - 1.00 kN 2.00 m - 2.00 kN 1.00 m



discontinuities in the shear diagram. The load of2.00 kN causes the shear diagram to jump DOWN by 2.00 kN at x =

1.00 m. Just slightly to the left ofx = 1.00 m, the shear force is -1.00 kN. Adding -2.00 kN to this value gives a

shear force of-3.00 kN just slightly to the right ofx = 1.00 m.


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Sum of Vertical Forces(Reaction Ay assumed positive upward)

Ay= 2.40 kN + 1.20 kN - 1.50 kN

Sum of Moments about A (Right hand rule sign convention)(Reaction MA assumed positive clockwise)

MA= - 2.40 kN 2.00 m - 1.20 kN 4.00 m + 1.50 kN 3.00 m


At x = 3.00 m, a concentrated load of1.50 kN acts upward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of1.50 kN causes the shear diagram to jump UP by 1.50 kN at x =

3.00 m. Just slightly to the left ofx = 3.00 m, the shear force is -0.3000 kN. Adding +1.50 kN to this value gives ashear force of1.20 kN just slightly to the right ofx = 3.00 m.


discontinuities in the shear diagram. The load of1.20 kN causes the shear diagram to jump DOWN by 1.20 kN at x =

4.00 m. Just slightly to the left ofx = 4.00 m, the shear force is 1.20 kN. Adding -1.20 kN to this value gives a shear

force of0.00 kN just slightly to the right ofx = 4.00 m.


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Ay+ By= 100.00 lb/ft 4.00 ft + 200.00 lb/ft 4.00 ft

Sum of Moments about A (Right hand rule sign convention)(8.00 ft) By= 100.00 lb/ft 4.00 ft 2.00 ft + 200.00 lb/ft 4.00 ft 6.00 ft

The total magnitude of the distributed load in this region is 100.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 4.00 ft is -400.00 lb (i.e., -100.00 lb/ft 4.00 ft ). The change in shearbetween x = 0.00 ft and x = 4.00 ft is equal to the area under the load diagram between these two points. At x =0.00 ft, the shear force is 500.00 lb. Adding -400.00 lb to this value gives a shear force of V = 100.00 lb at x = 4.00ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -100.00 lb/ft). Since itsslope is constant, the shear curve is linear in this region.

The total magnitude of the distributed load in this region is 200.00 lb/ft, acting DOWNWARD. The area

under the load diagram between x = 4.00 ft and x = 8.00 ft is -800.00 lb (i.e., -200.00 lb/ft 4.00 ft ). The change

in shear between x = 4.00 ft and x = 8.00 ft is equal to the area under the load diagram between these two points.

At x = 4.00 ft, the shear force is 100.00 lb. Adding -800.00 lb to this value gives a shear force of V = -700.00 lb at x

= 8.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -200.00 lb/ft).

Since its slope is constant, the shear curve is linear in this region.


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Ay+ By= 15.00 kN/m 2.00 m

Sum of Moments about A (Right hand rule sign convention)(4.50 m) By= 15.00 kN/m 2.00 m 1.50 m

The total magnitude of the distributed load in this region is 15.00 kN/m, acting DOWNWARD. The area under the

load diagram between x = 0.50 m and x = 2.50 m is -30.00 kN (i.e., -15.00 kN/m 2.00 m ). The change in shear

between x = 0.50 m and x = 2.50 m is equal to the area under the load diagram between these two points. At x =

0.50 m, the shear force is 20.00 kN. Adding -30.00 kN to this value gives a shear force of V = -10.00 kN at x = 2.50

m. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -15.00 kN/m). Since its

slope is constant, the shear curve is linear in this region.


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Solucion A) EQUILIBRIUM EQUATIONS


Ay+ By= 2.00 kN/m 2.00 m

Sum of Moments about A (Right hand rule sign convention)(4.00 m) By= 2.00 kN/m 2.00 m 2.00 m + 4.00 kN-m

The total magnitude of the distributed load in this region is 2.00 kN/m, acting DOWNWARD. The area under theload diagram between x = 2.00 m and x = 3.00 m is -2.00 kN (i.e., -2.00 kN/m 1.00 m ). The change in shearbetween x = 2.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x =2.00 m, the shear force is -1.00 kN. Adding -2.00 kN to this value gives a shear force of V = -3.00 kN at x = 3.00 m.The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -2.00 kN/m). Since itsslope is constant, the shear curve is linear in this region.

A concentrated moment of4.00 kN-m acts clockwise on the beam at x = 2.00 m. While concentrated

moments don't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise

concentrated moment of4.00 kN-m causes the moment diagram to jump UP by -4.00 kN-m at x = 2.00 m. Just

slightly to the left ofx = 2.00 m, the bending moment is 1.00 kN-m. Adding +4.00 kN-m to this value gives a

bending moment of5.00 kN-m just slightly to the right ofx = 2.00 m.


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Ay+ By= 800.00 lb/ft 12.00 ft


The total magnitude of the distributed load in this region is 800.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 12.00 ft is -9,600.00 lb (i.e., -800.00 lb/ft 12.00 ft ). The change inshear between x = 0.00 ft and x = 12.00 ft is equal to the area under the load diagram between these two points. Atx = 0.00 ft, the shear force is 5,623.53 lb. Adding -9,600.00 lb to this value gives a shear force of V = -3,976.47 lbat x = 12.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -800.00lb/ft). Since its slope is constant, the shear curve is linear in this region.

A concentrated moment of10,000.00 lb-ft acts clockwise on the beam at x = 14.00 ft. While concentrated momentsdon't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise concentratedmoment of10,000.00 lb-ft causes the moment diagram to jump UP by -10,000.00 lb-ft at x = 14.00 ft. Just slightly

to the left ofx = 14.00 ft, the bending moment is 1,929.42 lb-ft. Adding +10,000.00 lb-ft to this value gives abending moment of11,929.41 lb-ft just slightly to the right ofx = 14.00 ft. Solucion: A) EQUILIBRIUMEQUATIONS


Ay+ By= 800.00 lb/ft 12.00 ft


The total magnitude of the distributed load in this region is 800.00 lb/ft, acting DOWNWARD. The area under theload diagram between x = 0.00 ft and x = 12.00 ft is -9,600.00 lb (i.e., -800.00 lb/ft 12.00 ft ). The change inshear between x = 0.00 ft and x = 12.00 ft is equal to the area under the load diagram between these two points. At

x = 0.00 ft, the shear force is 5,623.53 lb. Adding -9,600.00 lb to this value gives a shear force of V = -3,976.47 lbat x = 12.00 ft. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -800.00lb/ft). Since its slope is constant, the shear curve is linear in this region.

A concentrated moment of10,000.00 lb-ft acts clockwise on the beam at x = 14.00 ft. While concentrated moments

don't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise concentrated

moment of10,000.00 lb-ft causes the moment diagram to jump UP by -10,000.00 lb-ft at x = 14.00 ft. Just slightly

to the left ofx = 14.00 ft, the bending moment is 1,929.42 lb-ft. Adding +10,000.00 lb-ft to this value gives a

bending moment of11,929.41 lb-ft just slightly to the right ofx = 14.00 ft.


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Ay+ By= 20.00 kN/m 2.00 m + 70.00 kN

Sum of Moments about A (Right hand rule sign convention)

(3.00 m) By= 20.00 kN/m 2.00 m 2.00 m + 70.00 kN 1.00 m

The total magnitude of the distributed load in this region is 20.00 kN/m, acting DOWNWARD. The area under theload diagram between x = 1.00 m and x = 3.00 m is -40.00 kN (i.e., -20.00 kN/m 2.00 m ). The change in shearbetween x = 1.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x =1.00 m, the shear force is -10.00 kN. Adding -40.00 kN to this value gives a shear force of V = -50.00 kN at x = 3.00m. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -20.00 kN/m). Since itsslope is constant, the shear curve is linear in this region.


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x = 1.00 m. Just slightly to the left ofx = 1.00 m, the shear force is 60.00 kN. Adding -70.00 kN to this value gives

a shear force of-10.00 kN just slightly to the right ofx = 1.00 m.


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Ay+ By= 3,500.00 lb

Sum of Moments about A (Right hand rule sign convention)(16.00 ft) By= 3,500.00 lb 14.00 ft + 4,000.00 lb-ft

At x = 16.00 ft, a concentrated load of3,500.00 lb acts downward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of3,500.00 lb causes the shear diagram to jump DOWN by 3,500.00lb at x = 16.00 ft. Just slightly to the left ofx = 16.00 ft, the shear force is 187.50 lb. Adding -3,500.00 lb to thisvalue gives a shear force of-3,312.50 lb just slightly to the right ofx = 16.00 ft.

A concentrated moment of4,000.00 lb-ft acts clockwise on the beam at x = 0.00 ft. While concentrated momentsdon't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise concentratedmoment of4,000.00 lb-ft causes the moment diagram to jump UP by -4,000.00 lb-ft at x = 0.00 ft. Just slightly tothe left ofx = 0.00 ft, the bending moment is 0.00 lb-ft. Adding +4,000.00 lb-ft to this value gives a bendingmoment of4,000.00 lb-ft just slightly to the right ofx = 0.00 ft.


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Ay+ By= 0.5(1,000.00 lb/ft) 12.00 ft + 0.5(1,000.00 lb/ft) 12.00 ft

Sum of Moments about A (Right hand rule sign convention)(24.00 ft) By= 0.5(1,000.00 lb/ft)(12.00 ft)(8.00 ft) + (1,000.00 lb/ft)(12.00 ft)(18.00 ft) + 0.5(-1,000.00

lb/ft)(12.00 ft)(20.00 ft)

The area under the load diagram between x = 0.00 ft and x = 12.00 ft is -6,000.00 lb. The change in shear betweenx = 0.00 ft and x = 12.00 ft is equal to the area under the load diagram between these two points. At x = 0.00 ft,the shear force is 6,000.00 lb. Adding -6,000.00 lb to this value gives a shear force of V = 833.33E-06 lb at x =12.00 ft. The shear curve is parabolic in this region.

The area under the load diagram between x = 12.00 ft and x = 24.00 ft is -6,000.00 lb. The change in shear between

x = 12.00 ft and x = 24.00 ft is equal to the area under the load diagram between these two points. At x = 12.00 ft,

the shear force is -833.33E-06 lb. Adding -6,000.00 lb to this value gives a shear force of V = -6,000.00 lb at x =

24.00 ft. The shear curve is parabolic in this region.


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Ay+ By= 10.00 kN + 5.00 kN + 15.00 kN

Sum of Moments about A (Right hand rule sign convention)(5.00 m) By= 10.00 kN 1.00 m + 5.00 kN 2.00 m + 15.00 kN 3.00 m

At x = 1.00 m, a concentrated load of10.00 kN acts downward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of10.00 kN causes the shear diagram to jump DOWN by 10.00 kN atx = 1.00 m. Just slightly to the left ofx = 1.00 m, the shear force is 17.00 kN. Adding -10.00 kN to this value givesa shear force of7.00 kN just slightly to the right ofx = 1.00 m.

At x = 2.00 m, a concentrated load of5.00 kN acts downward on the beam. Concentrated forces creatediscontinuities in the shear diagram. The load of5.00 kN causes the shear diagram to jump DOWN by 5.00 kN at x =


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2.00 m. Just slightly to the left ofx = 2.00 m, the shear force is 7.00 kN. Adding -5.00 kN to this value gives a shearforce of2.00 kN just slightly to the right ofx = 2.00 m.



x = 3.00 m. Just slightly to the left ofx = 3.00 m, the shear force is 2.00 kN. Adding -15.00 kN to this value gives ashear force of-13.00 kN just slightly to the right ofx = 3.00 m.


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Ay+ By= 2.00 kN/m 2.00 m

Sum of Moments about A (Right hand rule sign convention)

(4.00 m) By= 2.00 kN/m 2.00 m 2.00 m

The total magnitude of the distributed load in this region is 2.00 kN/m, acting DOWNWARD. The area under the load

diagram between x = 1.00 m and x = 3.00 m is -4.00 kN (i.e., -2.00 kN/m 2.00 m ). The change in shear between

x = 1.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x = 1.00 m,

the shear force is 2.00 kN. Adding -4.00 kN to this value gives a shear force of V = -2.00 kN at x = 3.00 m. The

slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -2.00 kN/m). Since its slope is

constant, the shear curve is linear in this region.


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Ay+ By= 0.5(200.00 lb/ft) 20.00 ft

Sum of Moments about A (Right hand rule sign convention)(20.00 ft) By= 0.5(200.00 lb/ft)(20.00 ft)(23.333 ft)

The area under the load diagram between x = 20.00 ft and x = 30.00 ft is -1,500.00 lb. The change in shear between

x = 20.00 ft and x = 30.00 ft is equal to the area under the load diagram between these two points. At x = 20.00 ft,

the shear force is 1,500.00 lb. Adding -1,500.00 lb to this value gives a shear force of V = 166.67E-06 lb at x =

30.00 ft. The shear curve is parabolic in this region.


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Ay+ By= 0.5(1,000.00 N/m) 3.00 m

Sum of Moments about A (Right hand rule sign convention)(2.00 m) By= 0.5(1,000.00 N/m)(3.00 m)(1.00 m)

The area under the load diagram between x = 1.00 m and x = 3.00 m is -1,333.33 N. The change in shear between x= 1.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x = 1.00 m, theshear force is 583.33 N. Adding -1,333.33 N to this value gives a shear force of V = -750.00 N at x = 3.00 m. Theshear curve is parabolic in this region.


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Solucion A) EQUILIBRIUM EQUATIONS


Ay+ By= + 30.00 lb/ft 10.00 ft

Sum of Moments about A (Right hand rule sign convention)(40.00 ft) By= 400.00 lb-ft + 30.00 lb/ft 10.00 ft 45.00 f

A concentrated moment of400.00 lb-ft acts clockwise on the beam at x = 20.00 ft. While concentrated momentsdon't directly affect the shear diagram, they cause discontinuities in the moment curve. The clockwise concentratedmoment of400.00 lb-ft causes the moment diagram to jump UP by -400.00 lb-ft at x = 20.00 ft. Just slightly to theleft ofx = 20.00 ft, the bending moment is -950.00 lb-ft. Adding +400.00 lb-ft to this value gives a bending momentof-550.00 lb-ft just slightly to the right ofx = 20.00 ft.

The total magnitude of the distributed load in this region is 30.00 lb/ft, acting DOWNWARD. The area under the loaddiagram between x = 40.00 ft and x = 50.00 ft is -300.00 lb (i.e., -30.00 lb/ft 10.00 ft ). The change in shearbetween x = 40.00 ft and x = 50.00 ft is equal to the area under the load diagram between these two points. At x =40.00 ft, the shear force is 300.00 lb. Adding -300.00 lb to this value gives a shear force of V = 25.00E-06 lb at x =

50.00 ft.

The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -30.00 lb/ft). Since

its slope is constant, the shear curve is linear in this region.


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Sum of Vertical Forces

(Reactions Ay and By assumed positive upward)Ay+ By= + 8.00 kN/m 3.00 m

Sum of Moments about A (Right hand rule sign convention)(2.00 m) By= - 2.25 kN-m + 8.00 kN/m 3.00 m 0.50 m

A concentrated moment of2.25 kN-m acts counterclockwise on the beam at x = 0.00 m. While concentratedmoments don't directly affect the shear diagram, they cause discontinuities in the moment curve. Thecounterclockwise concentrated moment of2.25 kN-m causes the moment diagram to jump DOWN by -2.25 kN-m atx = 0.00 m. Just slightly to the left ofx = 0.00 m, the bending moment is 0.00 kN-m. Adding -2.25 kN-m to thisvalue gives a bending moment of-2.25 kN-m just slightly to the right ofx = 0.00 m.

The total magnitude of the distributed load in this region is 8.00 kN/m, acting DOWNWARD. The area under the load

diagram between x = 1.00 m and x = 3.00 m is -16.00 kN (i.e., -8.00 kN/m 2.00 m ). The change in shear

between x = 1.00 m and x = 3.00 m is equal to the area under the load diagram between these two points. At x =

1.00 m, the shear force is 11.12 kN. Adding -16.00 kN to this value gives a shear force of V = -4.87 kN at x = 3.00

m. The slope of the shear curve is equal to the magnitude of the distributed load w (i.e., w = -8.00 kN/m). Since its

slope is constant, the shear curve is linear in this region.


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08-parcial 1 (shaums)

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