bending moment

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Shear Forces and Bending Moments

Problem 4.3-1

Problem 4.3-1 Calculate the shear forceCalculate the shear forceV V and bending momentand bending moment M M at a cross section just to the left of the 1600-lb load acting on the simple at a cross section just to the left of the 1600-lb load acting on the simple beam

beam AB ABshown in the figure.shown in the figure.

Solution 4.3-1

Solution 4.3-1 Simple beamSimple beam

4

Shear Forces and

Bending Moments

259 259 A A BB 1600 lb 1600 lb 800 lb 800 lb 120 in. 120 in. 3 30 0 iinn.. 660 0 iinn.. 330 0 iinn..  _M_M_A_A_0:_0: _R_R_B_B₁₄₀₁₄₀₀₀_lb_lb  _M_M_B_B_0:_0: _R_R_A_A₁₀₀₁₀₀₀₀_lb_lb

Free-body diagram of segment

Free-body diagram of segment DB DB

 42,00042,000lb-in.lb-in. © ©_M_M_D_D_0:_0: _M_M₍₁₄₀₀₍₁₄₀₀_lb)(30_lb)(30_in.)_in.)  200200lblb © ©_F_F_VERT_VERT0:0: V V 16001600lblb14001400lblb A A BB 1600 lb 1600 lb 800 lb 800 lb 3 30 0 iinn.. 660 0 iinn.. 330 0 iinn.. D D R R_A_A R R_B_B B B 1600 lb 1600 lb 30 in. 30 in. D D R R_B_B V V M M Problem 4.3-2

Problem 4.3-2 Determine the shear forceDetermine the shear force V V and bending momentand bending moment M M at the midpoint

at the midpointC C of the simple beamof the simple beam AB ABshown in the figure.shown in the figure.

Solution 4.3-2

Solution 4.3-2 Simple beamSimple beam

A A C C B B 2.0 kN/m 2.0 kN/m 6.0 kN 6.0 kN 1 1..0 0 mm 11..0 0 mm 4.0 m 4.0 m 2.0 m 2.0 m A A C C B B 2.0 kN/m 2.0 kN/m 6.0 kN 6.0 kN 1 1..0 0 mm 11..0 0 mm 22..0 0 mm R R_A_A R R_B_B   M M _A_A0:0: R R_B_B4.5kN4.5kN  _M_M_B_B_0:_0: _R_R_A_A_5.5kN_5.5kN

Free-body diagram of segment AC AC

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Problem 4.3-3

Problem 4.3-3 Determine the shear forceDetermine the shear force V V and bending momentand bending moment M M atat the midpoint of the beam with overhangs (see figure). Note that one load the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward.

acts downward and the other upward.

Solution 4.3-3

Solution 4.3-3 Beam with overhangsBeam with overhangs

P P P P b b b b LL  PP

¢

1122bb L L

≤

(upward)(upward) R R A A 1 1 L L[[PP(( L Lbbbb))]] © ©_M_M_B_B00 Free-body diagram

Free-body diagram ((C C is the midpointis the midpoint))

M M PLPL 2 2 PbPbPbPb PL PL 2 2 00 M M _P_P

¢

₁₁22bb L L

≤

¢

L L 2 2

≤

PP

¢

bb L L 2 2

≤

¢

₁₁22bb L L

≤

PP © ©_F_F_VERT_VERT_0:_0: © ©_M_M_A_A0:0: R R_B_BPP

¢

1122bb L L

≤

(downward)(downward) P P P P b b b b LL A A BB R R_A_A _R_R B B P P b b LL//22 A A C C R R_A_A _V_V M M Problem 4.3-4

Problem 4.3-4 Calculate the shear forceCalculate the shear forceV V and bending momentand bending moment M M at aat a cross section located 0.5 m from the fixed support of the cantilever beam cross section located 0.5 m from the fixed support of the cantilever beam AB

ABshown in the figure.shown in the figure.

Solution 4.3-4

Solution 4.3-4 CantiCantileverleverbeambeam

A A B B 1.5 kN/m 1.5 kN/m 4.0 kN 4.0 kN 1.0 m 1.0 m 1 1..00mm 22..00mm

Free-body diagram of segment DB DB Point

Point D Dis 0.5is 0.5m from m from supportsupport A A.. 9.59.5kNkN mm

 _2.0_2.0_kN_kN mm7.57.5kNkNmm  (1.5(1.5kNkN



m)(2.0m)(2.0 m)(2.5m)(2.5m)m) © ©_M_M_D_D0:0: M M (4.0(4.0 kN)(0.5kN)(0.5m)m)  4.04.0kNkN3.03.0kNkN7.07.0kNkN V V 4.04.0kNkN(1.5(1.5kNkN



m)(2.0m)(2.0m)m) © ©_F_F_VERT_VERT0:0: A A B B 1.5 kN/m 1.5 kN/m 4.0 kN 4.0 kN 1.0 m 1.0 m 1 1..00mm 22..00mm D D B B 1.5 kN/m 1.5 kN/m 4.0 kN 4.0 kN 1.0 m 1.0 m 0.5 m 0.5 m 2.0 m 2.0 m V V M M

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Problem 4.3-5 Determine the shear force V and bending moment M at a cross section located 16 ft from the left-hand end A of the beam with an overhang shown in the figure.

Solution 4.3-5 Beam with an overhang

SECTION 4.3 Shear Forces and Bending Moments

261

A B C 400 lb/ft 200 lb/ft 6 ft 6 ft 10 ft 10 ft _M_B_{0: R}_A_{2460 lb} _M_A_{0: R}_B_{2740 lb}

Free-body diagram of segment AD

Point D is 16 ft from support A.

 _{4640 lb-ft} _{(400 lb}



_{ft)(10 ft)(11 ft)} ©_M_D_{0: M}_{(2460 lb)(16 ft)}  _{1540 lb} V 2460 lb(400 lb



ft)(10 ft) ©_F_VERT_0: A B C 400 lb/ft 200 lb/ft 6 ft 6 ft 10 ft 10 ft R_A R_B A D 400 lb/ft 6 ft 10 ft R A V M

Problem 4.3-6 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C . The loads consist of a horizontal force P₁ 4.0 kN acting at the end of a vertical arm and a vertical force P₂8.0 kN acting at the end of the overhang.

Determine the shear force V and bending moment M at a cross section located 3.0 m from the left-hand support.

( Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.3-6 Beam with vertical arm

4.0m 1.0m B A C P₂= 8.0 kN P₁= 4.0 kN 1.0 m 4.0m 1.0m B A P₂= 8.0 kN P₁= 4.0 kN 1.0 m R_A R_B _M_B_0: _R_A_{1.0 kN (downward)} _M A0: R B9.0 kN (upward)

Free-body diagram of segment AD Point D is 3.0 m from support A.

 _{7.0 kN} m ©_M_D0: M   R_A(3.0 m)4.0 kN  m ©_F_VERT_0: _V _R_A _{1.0 kN} 3.0 m A D R_A V M 4.0kN •m

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Problem 4.3-7 The beam ABCD shown in the figure has overhangs at each end and carries a uniform load of intensity q.

For what ratio b/L will the bending moment at the midpoint of the beam be zero?

Solution 4.3-7 Beam with overhangs

q b b L D A B C

From symmetry and equilibrium of vertical forces: R_B  R_C  q

¢

b L

2

≤

Free-body diagram of left-hand half of beam: Point E is at the midpoint of the beam.

Solve for b / L : b L 1 2 _q

¢

_b L 2

≤ ¢

L 2

≤

q

¢

1 2

≤ ¢

b L 2

≤

2 ₀  R B

¢

L 2

≤

q

¢

1 2

≤ ¢

b L 2

≤

2 0 ©_M_E  ₀ _{ } q b b L D A B C R_B R_C q b L/2 A R_B V M =0 (Given) E

Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the bowstring of the bow shown in the figure. Determine the bending moment at the midpoint of the bow.

350 mm

1400 mm 70°

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Solution 4.3-8 Archer’s bow

263

P130N _70° H _{1400 mm} _1.4m b_{350 mm} 0.35m

Free-body diagram of point A

T _{tensile force in the bowstring} _F_HORIZ_0: _2T cos _P₀

T  P

2 cos b

Free-body diagram of segment BC

Substitute numerical values:

M _{108 N} m M 130 N 2

B

1.4 m 2 (0.35 m)(tan 70)

R

P 2

¢

H 2 b tan b

≤

M  _T

¢

H 2 cos b  bsin b

≤

T (cos b)

¢

H 2

≤

 T (sin b) (b) M 0 ©_M_C 0 _{ } b H P A  C B P A T  T H 2 C b B T  M

(6)

Problem 4.3-9 A curved bar ABC is subjected to loads in the form of two equal and opposite forces P, as shown in the figure. The axis of the bar forms a semicircle of radius r .

Determine the axial force N , shear force V , and bending moment M acting at a cross section defined by the angle .

Solution 4.3-9 Curved bar

P P P C B A O r A V N M   M _Nr_{Pr sin u} ©_M_O₀ _{ } _M_Nr₀ V _{P cos u}



F V  0  R a  V  _{P cos u}  ₀ N  _{P sin u} ©_F_N  0 Q b  N _Psin u0 P P P C B A O r A V N M   O Pcos  Psin  B

Problem 4.3-10 Under cruising conditions the distributed load acting on the wing of a small airplane has the idealized variation shown in the figure.

Calculate the shear force V and bending moment M at the inboard end of the wing.

Solution 4.3-10 Airplane wing

1.0 m 1600 N/m _{900 N/m} 2.6 m 2.6 m 1.0 m 1600 N/m _{900 N/m} 2.6 m 2.6 m A B V M Shear Force F _VERT0 c T 

(Minus means the shear force acts opposite to the direction shown in the figure.)

V  _{6040 N} _{6.04 kN} 1 2 (900 N



m)(1.0 m)0 V 1 2(700 N



m)(2.6 m)(900 N



m)(5.2 m) Bending Moment M 788.67 N•m 12,168 N•m2490N •m _{15,450 N}•m _{15.45 kN} m 1 2(900 N



m)(1.0 m)

¢

5.2 m 1.0 m 3

≤

0 (900 N



m)(5.2 m)(2.6 m)  M 1 2 (700 N



m)(2.6 m)

¢

2.6 m 3

≤

©_M_A0 _ A B 3 2 1 700 N/m 900 N/m

(7)

Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E . One end of the cable is attached to the beam at point B.

What is the force P in the cable if the bending moment in the beam just to the left of point C is equal numerically to 640 lb-ft? ( Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.3-11 Beam with a cable

265

A E P C D B Cable 8 ft 6ft 6ft 6ft UNITS: P in lb M in lb-ft

Free-body diagram of section AC

Numerical value of M equals 640 lb-ft.

and P1200 lb ∴640 lb-ft8P 15 lb-ft M  8P 15 lb-ft M 4P 5 (6 ft) 4P 9 (12 ft)0 ©_M_C0 _ A E P C D B Cable 8 ft 6ft 6ft 6ft P __ 9 __ 9 4P ₄_P A P C B 6ft 6ft P __ 5 __ 5 N M V __ 9 4P 4P 3P

Problem 4.3-12 A simply supported beam AB supports a trapezoidally distributed load (see figure). The intensity of the load varies linearly from 50 kN/m at support A to 30 kN/m at support B.

Calculate the shear force V and bending moment M at the midpoint

of the beam. A B

50 kN/m

30 kN/m

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Solution 4.3-12 Beam with trapezoidal load Reactions R A65kN R_B55kN _{30 kN}



_{m)(3 m)} ₀ R A R B1



2(50 kN



m ©_F_VERT  ₀c _{(20 kN}



_{m)(3 m) (}1

_

2) (2 m)0 ©_M_B₀__R_A_{(3 m)}_{(30 kN}



_{m)(3 m)(1.5 m)}

Free-body diagram of section CB Point C is at the midpoint of the beam.

F _VERT0 c_ T _55kN₀  M (30 kN/m)(1.5 m)(0.75 m) _{(55kN)(1.5 m)}₀ M _{45.0 kN}m 1

_

2(10 kN



m)(1.5 m)(0.5 m) ©_M_C₀ _ V  _{2.5 kN} V (30 kN



m)(1.5 m)1 2(10 kN



m)(1.5 m) B A 50 kN/m 30 kN/m 3 m R_A _R B B V 40 kN/m 30 kN/m 1.5 m 55 kN C M

Problem 4.3-13 Beam ABCD represents a reinforced-concrete

foundation beam that supports a uniform load of intensity q₁3500 lb/ft (see figure). Assume that the soil pressure on the underside of the beam is uniformly distributed with intensity q₂.

(a) Find the shear force V Band bending moment M Bat point B.

(b) Find the shear force V mand bending moment M mat the midpoint

of the beam.

Solution 4.3-13 Foundation beam

A B C D 3.0ft 3.0ft q₂ q₁= 3500 lb/ft 8.0 ft _F_VERT_0: _q₂_(14ft)  _q₁_{(8 ft)}

(a) V and M at point B

_F_VERT_0: ©_M_B  _{0: M}_B  _{9000 lb-ft} V B  6000 lb ∴q2 8 14 q12000 lb



ft

(b) V and M at midpoint E

F _VERT 0: V _m (2000lb/ft)(7ft)  (3500lb/ft)(4 ft)  M _E _0: M m (2000lb/ft)(7ft)(3.5 ft) _{(3500 lb/ft)(4ft)(2 ft)} M m21,000 lb-ft V m 0 A B C D 3.0ft 3.0ft q₂ q₁= 3500 lb/ft 8.0 ft A B 3 ft 2000 lb/ft V _B M _B A B E 4 ft 2000 lb/ft 3500 lb/ft 3 ft M _m V _m

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Problem 4.3-14 The simply-supported beam ABCD is loaded by a weight W 27 kN through the arrangement shown in the figure. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm.

Calculate the axial force N , shear force V , and bending moment M at section C, which is just to the left of the vertical arm.

( Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.3-14 Beam with cable and weight

267

A E D C B W = 27 kN 2.0 m 2.0 m 2.0 m Cable 1.5 m R_A_18kN _R D9 kN

Free-body diagram of pulley at B

A E D C B 27 kN 2.0 m 2.0 m 2.0 m Cable _{1.5 m} R_A R_D 27 kN 21.6 kN 10.8 kN 27 kN

Free-body diagram of segment ABC of beam

©_M_C0: M 50.4 kN m ©_F_VERT0: V 7.2 kN ©_F_HORIZ0: N 21.6 kN (compression) A N M C B 21.6 kN 2.0 m 2.0 m V 10.8 kN 18 kN

(10)

Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) with an angular acceleration . Each of the two arms has weight w per unit length and supports a weight W 2.0 wL at its end.

Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b  L /9 and c L /10.

Solution 4.3-15 Rotating centrifuge

b c L W x W y  b c L x W g __ ₍_L₊_b₊_c₎_ w x g

__

Tangential acceleration  _r 

Maximum V and M occur at x b.

w L 2_ 6g (2 L3b) W  g ( Lbc) ( Lc) 



Lb b w g x( x  b)dx M max W  g ( Lbc) ( Lc) wL 2g ( L2b) W  g ( Lbc) V max W g( Lbc)



Lb b w g x dx Inertial force Mr   W g r 

Substitute numerical data:

M max  229wL3 75g V max  91wL2 30g W _{2.0 wL} _b L 9 c L 10

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Shear-Force and Bending-Moment Diagrams

When solving the problems for Section 4.5, draw the shear-force and bending-moment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values.

Probs. 4.5-1 through 4.5-10 are symbolic problems and Probs. 4.5-11 through 4.5-24 are numerical problems. The remaining problems (4.5-25 through 4.5-30) involve specialized topics, such as optimization, beams with hinges, and moving loads.

Problem 4.5-1 Draw the shear-force and bending-moment diagrams for a simple beam AB supporting two equal concentrated loads P (see figure).

Solution 4.5-1 Simple beam

SECTION 4.5 Shear-Force and Bending-Moment Diagrams

269

A B L P P a a A B L P P a a R_A=P R_B=P P _P V Pa M 0 0

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Problem 4.5-2 A simple beam AB is subjected to a counterclockwise couple of moment M ₀acting at distance a from the left-hand support (see figure).

Draw the shear-force and bending-moment diagrams for this beam.

A B L a M ₀ A B L a M ₀ M0 L 0 V M M ₀a L 0  M ₀(1 a L ) R_A= M 0 L R B= M ₀ L

Problem 4.5-3 Draw the shear-force and bending-moment diagrams for a cantilever beam AB carrying a uniform load of intensity q over one-half of its length (see figure).

Solution 4.5-3 Cantilever beam

A B q L — 2 L_— 2 A B q L — 2 L — 2 qL — 2 V M qL2 0 0 M _A= 3qL2 8 R_A= qL 2 3qL2 8 8  

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Problem 4.5-4 The cantilever beam AB shown in the figure is subjected to a concentrated load P at the midpoint and a counterclockwise couple of moment M ₁PL /4 at the free end.

271

A B P L — 2 L — 2 M ₁= —–PL 4 M _A P R_A L/2 L/2 A B M ₁ PL 4 M _APL 4 R_AP V M 0 0 PL 4 PL 4 P

Problem 4.5-5 The simple beam AB shown in the figure is subjected to a concentrated load P and a clockwise couple M ₁_PL_{/4 acting at the}

third points.

A B P L — 3 L — 3 L — 3 M ₁= —–PL 4 A B P L — 3 L — 3 L — 3 M ₁= —–PL 4 R_A= —–5P 12 R B= 7P —– 12 5P /12 V M 0 0 5PL /36 7PL /36 _PL_/18 ₇_P_/12

(14)

Problem 4.5-6 A simple beam AB subjected to clockwise couples M 1

and 2 M ₁acting at the third points is shown in the figure.

A B M ₁ 2 M ₁ L — 3 L — 3 L — 3 A B M ₁ 2 M ₁ L — 3 L — 3 L — 3 R B= 3 M ₁ —– L R_A= 3—– M 1 L V 3—– M 1 L 0 M 0 M ₁  M ₁  M ₁

Problem 4.5-7 A simply supported beam ABC is loaded by a vertical load P acting at the end of a bracket BDE (see figure).

Draw the shear-force and bending-moment diagrams for beam ABC .

Solution 4.5-7 Beam with bracket

A C L D E P B L — 4 L — 4 L — 2 A C P B L — 4 —4 3 L R_A= —–P 2 RC = P —– 2 V M 0 0 P —– 2 PL —– 8 PL —– 4 3PL —– 8 P —– 2 

(15)

Problem 4.5-8 A beam ABC is simply supported at A and B and has an overhang BC (see figure). The beam is loaded by two forces P and a clockwise couple of moment Pa that act through the arrangement shown.

Draw the shear-force and bending-moment diagrams for beam ABC .

Solution 4.5-8 Beam with overhang

273

A C B a a a a P P _Pa C P P Pa a a a P P upper beam: B P P a a a 2P lower beam: C V 0 M 0 P Pa P

Problem 4.5-9 Beam ABCD is simply supported at B and C and has overhangs at each end (see figure). The span length is L and each overhang has length L /3. A uniform load of intensity q acts along the entire length of the beam.

q L L 3 D A B C L 3 q L L/ 3 –qL2_/₁₈ _–qL2_/₁₈ qL/ 3 L / 3 D A B C

__

5qL R_B= 6

__

qL – 3

__

qL – 2

__

5qL R_C= 6 V M X ₁

__

5qL2 72 0 0

__

qL 2 x1 L



5 6  0.3727 L

(16)

Problem 4.5-10 Draw the shear-force and bending-moment diagrams for a cantilever beam AB supporting a linearly varying load of maximum intensity q₀(see figure).

A B L q₀ A V M B L q₀ x

__

x q=q₀ L _q

__

0 L2 M _B= 6

__

q₀ x3 M = – 6 L

__

q₀ x2 V = – 2 L

__

q₀ L R_B= 2

__

q₀ L – 2

__

q₀ L2 – 6 0 0

Problem 4.5-11 The simple beam AB supports a uniform load of intensity q10 lb/in. acting over one-half of the span and a concentrated load P80 lb acting at midspan (see figure).

A B q= 10 lb/in. P= 80 lb = 40 in. L — 2 = 40 in. L — 2 A B 10 lb/in. P= 80 lb 40 in. 46 in. 6 in. 40 in. 60 R_B=340 lb R_A=140 lb 140 –340 V M M _max=5780 5600 (lb) (lb/in.) 0 0

(17)

Problem 4.5-12 The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length.

Solution 4.5-12 Beam with distributed loads

275

0.8 m 3000 N/m A B 0.8 m 1.6 m 0.8 m 3000 N/m A V M B 0.8 m 1.6 m 1500 N/m 1200 –1200 960 480 480 (N) (N

.

m) 0 0

Problem 4.5-13 A cantilever beam AB supports a couple and a concentrated load, as shown in the figure.

A B 5ft 5ft 200 lb 400 lb-ft A B 5ft 5ft 200 lb 400 lb-ft M _A=1600 lb-ft R_A=200 lb V M (lb) +200 –600 –1600 –1000 0 0 (lb-ft)

(18)

Problem 4.5-14 The cantilever beam AB shown in the figure is

subjected to a uniform load acting throughout one-half of its length and a concentrated load acting at the free end.

A B 2m 2m 2.5 kN 2.0 kN/m A B 2m 2m 2.5 kN 2.0 kN/m R_A= 6.5 kN M _A= 14 kN

.

m 6.5 –14.0 –5.0 2.5 V M (kN) (kN

.

m) 0 0

Problem 4.5-15 The uniformly loaded beam ABC has simple supports at A and B and an overhang BC (see figure).

A V M C B 72 in. 25 lb/in. 48 in. R_A= 500 lb R_B= 2500 lb 1200 500 20 in. –1300 –28,800 20 in. 40 in. (lb) (lb-in.) 0 0 5000 A C B 72 in. 25 lb/in. 48 in.

(19)

Problem 4.5-16 A beam ABC with an overhang at one end supports a uniform load of intensity 12 kN/m and a concentrated load of magnitude 2.4 kN (see figure).

277

A C B 1.6 m 1.6 m 1.6 m 2.4 kN 12 kN/m A C B 1.6 m 1.6 m 1.6 m 2.4 kN 2.4 13.2 5.76 –3.84 –6.0 12 kN/m V M M _max=7.26 R_A= 13.2 kN R_B= 8.4 kN (kN

.

m) 0 0 1.1m 1.1m 0.64 m M _max (kN)

Problem 4.5-17 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C . The loads consist of a horizontal force P₁400 lb acting at the end of the vertical arm and a vertical force P2900 lb acting at the

end of the overhang.

Draw the shear-force and bending-moment diagrams for this beam. ( Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)

Solution 4.5-17 Beam with vertical arm

4.0 ft 1.0 ft B A _C P₂= 900 lb P₁= 400 lb 1.0 ft V (lb) M (lb) 900 0 0 ₄₀₀ ₉₀₀ 4.0 ft 1.0 ft B A _C P₂= 900 lb P₁= 400 lb 1.0 ft R_A= 125 lb R_B= 1025 lb A 400 lb-ft 125 lb B 900 lb C 1025 lb ₁₂₅

(20)

Problem 4.5-18 A simple beam AB is loaded by two segments of uniform load and two horizontal forces acting at the ends of a vertical arm (see figure).

A B 4 kN/m 8 kN 4 kN/m 2 m 2 m 2m 2m 1 m 1 m 8 kN

Problem 4.5-19 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E . One end of the cable is attached to the beam at point B. The tensile force in the cable is 1800 lb.

Draw the shear-force and bending-moment diagrams for beam ABCD. ( Note: Disregard the widths of the beam and vertical arm and use center-line dimensions when making calculations.)

Solution 4.5-19 Beam with a cable

A E C D B Cable 8 ft 1800 lb 6ft 6ft 6ft A B 4 kN/m 4 kN/m 2 m 2 m 2m 2m R_A= 6 kN R_B= 10 kN V (kN) M (kN

.

m) 0 0 _2.0 16 kN

.

m 1.5 m 1.5 m 6.0 _10.0 4.5 4.0 16.0 12.0

Note: All forces have units of pounds. A E C D B Cable 8 ft 1800 lb 6ft 6ft 6ft 1800 lb R_D= 800 lb R_D= 800 lb

Free-body diagram of beam ABCD

A B C D 1800 1440 1800 1440 5760 lb-ft 800 1080 720 800 V (lb) M (lb-ft) 640 0 0 ₄₈₀₀ 4800 ₈₀₀ ₈₀₀ ₉₆₀

(21)

Problem 4.5-20 The beam ABCD shown in the figure has overhangs that extend in both directions for a distance of 4.2 m from the supports at B and C , which are 1.2 m apart.

Draw the shear-force and bending-moment diagrams for this overhanging beam.

279

A D 1.2 m 4.2m 4.2m 5.1 kN/m 5.1 kN/m 10.6 kN/m B C A D 1.2 m 4.2m 4.2m 5.1 kN/m 5.1 kN/m 10.6 kN/m B C R_B= 39.33 kN R_C= 39.33 kN V (kN) _32.97 6.36 0 32.97 _6.36 M 0 (kN

.

m) _61.15 _61.15 _59.24

Problem 4.5-21 The simple beam AB shown in the figure supports a concentrated load and a segment of uniform load.

A C B 2.0 k/ft 4.0 k 20 ft 10 ft 5 ft A C B 2.0 k/ft 4.0 k 10 ft 5 ft 5 ft R_A= 8 k R_B= 16 k M _max= 64 k-ft V (k) ₁₆ M (k-ft) 0 0 8 4 8 ft 12 ft 8 ft 12 ft 40 60 64 C C

(22)

Problem 4.5-24 A beam with simple supports is subjected to a

trapezoidally distributed load (see figure). The intensity of the load varies from 1.0 kN/m at support A to 3.0 kN/m at support B.

Draw the shear-force and bending-moment diagrams for this beam. Problem 4.5-22 The cantilever beam shown in the figure supports a concentrated load and a segment of uniform load.

Draw the shear-force and bending-moment diagrams for this cantilever beam.

A B 1.0 kN/m 3 kN 1.6 m 0.8 m 0.8 m A B 1.0 kN/m 3 kN 1.6 m 0.8 m 0.8 m R_A= 4.6 kN _6.24 M (kN

.

m) V (kN) 0 0 4.6 1.6 _2.56 _1.28 M _A= 6.24 kN

.

m

Problem 4.5-23 The simple beam ACB shown in the figure is subjected to a triangular load of maximum intensity 180 lb/ft.

B C A 180 lb/ft 7.0 ft 6.0 ft B C A 180 lb/ft 1.0 ft 6.0 ft R_A= 240 lb R_B= 390 lb M _max= 640 V (lb) ₃₀₀ M (lb-ft) 0 0 240 x₁= 4.0 ft ₃₉₀ 360 B A 3.0 kN/m 1.0 kN/m 2.4 m

(23)

281

B A 3.0 kN/m 1.0 kN/m 2.4 m R_A= 2.0 kN R_B= 2.8 kN Set V _{0: x}₁_1.2980m V  _2.0_x x 2 2.4 ( xmeters; V kN) M (kN

.

m) _2.8 2.0 0 0 M _max= 1.450 x₁= 1.2980 m x V (kN)

Problem 4.5-25 A beam of length L is being designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is qL2_/8.

However, if the supports of the beam are moved symmetrically toward the middle of the beam (as pictured), the maximum bending moment is reduced.

Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value.

Draw the shear-force and bending-moment diagrams for this condition.

Solution 4.5-25 Beam with overhangs Solution 4.5-24 Simple beam

A B L a q A B a q R_A=qL /2 R_B=qL /2 (L_a)/ 2 _(La)/ 2 M ₂ M ₁ M ₁ 0 M

The maximum bending moment is smallest when M ₁ M ₂(numerically). M 1 M 2 ( La)2 L(2a L) M ₂ R A

¢

a 2

≤

 qL2 8  qL 8 (2a L) M 1 q( L_a)2 8 0.2071 L 0.2071qL 0.02145qL2 0.2929 L _0.2071qL _0.2929_qL V 0 M 0 _0.02145qL2 _0.02145qL2 x₁ x₁ 0.2929qL x₁= 0.3536a = 0.2071 L qL 2 8 (32



2) 0.02145qL 2 M 1 M 2 q 8 ( La) 2 Solve for a: a(2



2) L0.5858 L

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Problem 4.5-26 The compound beam ABCDE shown in the figure consists of two beams ( AD and DE ) joined by a hinged connection at D. The hinge can transmit a shear force but not a bending moment. The loads on the beam consist of a 4-kN force at the end of a bracket attached at point B and a 2-kN force at the midpoint of beam DE .

Draw the shear-force and bending-moment diagrams for this compound beam.

Solution 4.5-26 Compound beam

A B C D E 4 kN 2 m 2 m 2 m 2 m 1 m 2 kN 1 m A B C D E 4 kN 1 m 1 m 1 m1 m 2 m 2 m 2 m 2 kN R_A= 2.5 kN _1.0 M (kN

.

m) 4 kN

.

m Hinge R_C= 2.5 kN R_E= 1 kN V (kN) ₀ 2.5 1.0 _1.5 D D 1.0 5.0 _2.0 0 2.67 m 1.0

Problem 4.5-27 The compound beam ABCDE shown in the figure consists of two beams ( AD and DE ) joined by a hinged connection at D. The hinge can transmit a shear force but not a bending moment. A force P acts upward at A and a uniform load of intensity q acts downward on beam DE .

Draw the shear-force and bending-moment diagrams for this compound beam.

Solution 4.5-27 Compound beam

A B E P C D 2 L L L L q A V M E B P C D 2 L L L L q PL D D P −P−qL −qL2 –qL L L R_C= P +2qL R_E= qL R_B=2P + qL 0 0 Hinge qL qL 2

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Problem 4.5-28 The shear-force diagram for a simple beam is shown in the figure.

Determine the loading on the beam and draw the bending-moment diagram, assuming that no couples act as loads on the beam.

Solution 4.5-28 Simple beam (V is given)

283

1.0 m 1.0 m 2.0 m 12 kN –12 kN 0 V 12 −12 12 0 0 V M 6.0 kN/m 12 kN A B 2m 1m 1 m (k N

.

m) (kN) R_A= 12kN R_B= 12kN

Problem 4.5-29 The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bending-moment diagram.

Solution 4.5-29 Forces on a beam (V is given)

4 ft 4ft 16ft 572 lb –128 lb 0 V 652 lb 500 lb 580 lb –448 lb 14.50 ft 572 2448 –2160 –128 0 0 V M 652 500 580 –448 (lb) (lb-ft) 4 ft 4ft 16ft 20 lb/ft 652 lb 700 lb 1028 lb 500 lb Force diagram

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Problem 4.5-30 A simple beam AB supports two connected wheel loads P and 2P that are distance d apart (see figure). The wheels may be placed at any distance x from the left-hand support of the beam.

(a) Determine the distance x that will produce the maximum shear force in the beam, and also determine the maximum shear force V _max. (b) Determine the distance x that will produce the maximum bending

moment in the beam, and also draw the corresponding bending-moment diagram. (Assume P_{10 kN, d}_{2.4 m, and L}_{12 m.)}

Solution 4.5-30 Moving loads on a beam

(a)Maximum shear force

By inspection, the maximum shear force occurs at support B when the larger load is placed close to, but not directly over, that support.

(b)Maximum bending moment

By inspection, the maximum bending moment occurs at point D, under the larger load 2P.

V max R BP

¢

3

d

L

≤

28 kN x_L_d_{9.6 m}

Reaction at support B:

Bending moment at D:

Eq.(1) Substitute x into Eq (1): R B  P 2

¢

3  d L

≤

 14 kN Note: R A  P 2

¢

3  d L

≤

 16 kN  PL 12

¢

3 d L

≤

2 78.4 kN m 

¢

L 6

≤ ¢

3 5d L

≤

2d ( Ld )

R

M _max  P L

B

3

¢

L 6

≤

2

¢

35d L

≤

2 (3 L5d ) Solve for x: x L 6

¢

3 5d L

≤

4.0 m dM D dx  P L (6 x3 L5d )0 P L[3 x 2__(3 L__5d ) x__2d ( L__d_{) ]}  P L (2d 3 x) ( L xd ) M D R B( L xd ) R_BP L x 2P L ( xd ) P L (2d 3 x) L B A x d P 2P L B A x d P 2P B A x = L−d P 2P R_B=P

(

3 −_Ld

)

R_A=_LPd d B A L P 2P x d D R_B 64 M max= 78.4 2.4 m 4.0m 5.6m 0 M (kN

.

m) P10kN d 2.4m L12 m