Problem 14.34 The load F = 4650 lb. Draw theshear force and bending moment diagrams for the beam.

Solution:Draw the FBD of the beam and determine the reactions at points Aand B.

ΣMA = 0 = −(4650 lb)(3 ft)+

3∫0

(3− x)(400x2)dx + By(5 ft)

0 = −13, 950 ft−lb +

3∫0

(1200x2 − 400x3)dx + By(5 ft)

By = 2250 lb ↑

ΣFy = 0 = −4650 lb + 2250 lb + Ay −3∫

0

(400x2)dx

Ay = 6000 lb ↑

Problem 14.35 If the load F = 2150 lb in Prob-lem 14.34, what are the maximum positive and negativevalues of the shear force and bending moment, and atwhat values of x do they occur?

Solution:Draw the FBD of the beam and determine the reactions at points Aand B.

ΣMA = 0 = −(2150 lb)(3 ft)+

3∫0

(3− x)(400x2)dx + By(5 ft)

0 = −6, 450 ft−lb +

3∫0

(1200x2 − 400x3)dx + By(5 ft)

By = 750 lb

ΣFy = 0 = −2150 lb + 750 lb + Ay −3∫

0

(400x2)dx

Ay = 5000 lb ↑We see that:

ANS: (V +)MAX = 1400 lb at 3 ft < x < 6 ft

ANS: (V−)MAX = −3600 lb at x = 3 ft

ANS: (M+)MAX = 1500 ft−lb at x = 6 ft

ANS: (M−)MAX = −2700 ft−lb at x = 3 ft

Problem 14.36 Draw the shear force and bending mo-ment diagrams.

Solution:Draw the FBD for the beam and determine the reactions at points Aand B.

ΣMA = 0 = (6, 000 N)(6 m) + 20, 000 N−m)

−(4, 000N)(6m)(3m)+By(6m)−[

12(4, 000 N/m)(6 m)

](8m)

By = 18, 667 N ↑

ΣFy = 0 = −6, 000N−(4, 000N/m)(6m)−[

12(4, 000 N/m)(6 m)

]+Ay+18, 667N

Ay = 23, 333 N ↑

Problem 7.101 Determine the Iy and ky .

x

y

h

b

Solution: Let dA = x dy = b dy. A =∫ h

0b dy = hb.

Iy =∫

Ax2 dA = h

∫ b

0x2 dx = h

[x3

3

]b

0=

b3h

3ky =

√Iy

A=

b√3

y

x

h

b

Problem 7.102 Determine Ix and kx by letting dA be(a) a horizontal strip of height dy; (b) a vertical strip ofwidth dx.

Solution: (See figure in Problem 7.101.)

(a) A =∫ h

0b dy = hb.

Ix =∫

Ay2 dA = b

∫ h

0y2 dy =

bh3

3,

kx =

√Ix

A=

h√3

(b) A =∫ b

0h dx = hb.

Ix = h

∫ b

0y2 dx = h

∫ h

0y2(

b

h

)dy =

bh3

3,

kx =

√Ix

A=

h√3

Problem 7.103 Determine Ixy .

Solution: (See figure in Problem 7.101.)

A =∫

AdA =

∫ b

0

∫ h

0dx dy = hb.

Ixy =∫

Axy dA =

∫ b

0

∫ h

0xy dx dy =

[y2

2

]h

0

[x2

2

]b

0=

h2b2

4

Problem 7.127 Determine Ix and kx for the compositearea by dividing it into rectangles 1 and 2 as shown, andcompare your results to those of Example 8.4.

x3 m

1 m

4 m1

2

y1 m

Solution:

Ix =∫

A1

y2 dA1 +∫

A2

y2 dA2

Ix =∫ 1

0

∫ 4

1y2 dy dx +

∫ 3

0

∫ 1

0y2 dy dx

Ix =∫ 1

0

[y3

3

]41

dx +∫ 3

0

[y3

3

]10

dx

Ix =∫ 1

0

[643− 1

3

]dx +

∫ 3

0

13

dx

Ix =633

x

]10

+13

x

]30

=633

+33

Ix = 21 + 1 = 22 ft4

Area = 3 + 3 = 6 ft2

kx =

√Ix

Area=

√226

= 1.91 ft

x

3 m

1 m

4 mA1

A2

y

1 m

3 m

Problem 7.128 Determine Iy and ky for the compositearea.

Solution:

Iy =∫

A1

x2 dA1 +∫

A2

x2 dA2

Iy =∫ 1

0

∫ 4

1x2 dy dx +

∫ 3

0

∫ 1

0x2 dy dx

Iy =∫ 1

0[x2y]41 dx +

∫ 3

0[x2y]10 dx

Iy =∫ 1

0(4x2 − x2) dx +

∫ 3

0x2 dx

=[

3x3

3

]10

+[

x3

3

]30

Iy = 1 + 9 = 10.00 m4

ky =

√106

= 1.29 m

x

3 m

1 m

4 mArea = 6m2

A1

A2

y

1 m

3 m

Problem 7.131 Determine Ix and kx.y

x

100mm

70mm

90 mm

30 mm

Solution:

Ix =∫

A1

y2 dA1 +∫

A2

y2 dA2 +∫

A3

y2 dA3

Ix =∫ −15

−45

∫ 70

0y2 dy dx +

∫ 45

−45

∫ 100

70y2 dy dx

+∫ 45

15

∫ 70

0y2 dy dx

Ix =∫ −15

−45

[y3

3

]700

dx +∫ 45

−45

[y3

3

]10070

dx

+∫ 45

15

[y3

3

]700

dx

Ix =∫ −15

−45

(703

3

)dx +

∫ 45

−45

(1003

3− 703

3

)dx

+∫ 45

15

(703

)3dx

Ix =703

3x

∣∣∣∣15

−45+(

1003

3− 703

3

)x

∣∣∣∣45

−45

+(

703

3

)x

∣∣∣∣45

−15

Ix =703

3(30) +

(1003

3− 703

3

)90 +

703

3(30)

Ix = 2.66× 107 mm4

kx =

√Ix

Area=

√2.66× 107

6900= 62.1 mm

kx = 62.1 mm

30 mm 30 mm

30 mm

70 mm

90 mm

A1

A2

A3

100 mm

x

y

Area = (100)(90)–(30)(70)

= 9000 – 2100Area = 6900 mm2

Problem 7.132 Determine Iy and ky .

Solution:

Iy =∫

A1

x2 dA1 +∫

A2

x2 dA2 +∫

A3

x2 dA3

Iy =∫ −15

−45

∫ 70

0x2 dy dx +

∫ 45

−45

∫ 100

70x2 dy dx

+∫ 45

15

∫ 70

0x2 dy dx

Iy =∫ −15

−45[x2y]700 dx +

∫ 45

−45(x2y)

∣∣∣∣100

70dx

+∫ 45

15(x2y)

∣∣∣∣70

0dx

Iy =∫ −15

−4570x2 dx +

∫ 45

−45(100− 70)x2 dx +

∫ 45

1570x2 dx

Iy = 70x3

3

∣∣∣∣−15

−45+ 30

x3

3

∣∣∣∣45

−45+ 70

x3

3

∣∣∣∣45

15

Iy = 70(

(−15)3

3− (−453)

3

)+ 30

(453

3− (−45)3

3

)

+ 70(

(45)3

3− (15)3

3

)

30 mm

30 mm

70 mm

30 mm

A1

A2

A3

100 mm

x

y

Area = 9000 mm2

– 2100 mm2

Area = 6900 mm2

Iy = 140(

453

3− 153

3

)+ 60

(453

3

)

Iy = 5.92× 106 mm4

ky =

√Iy

Area=

√5.92× 106

6900= 29.3 mm

ky = 29.3 mm

Problem 15.1 The beam consists of material withmodulus of elasticity E = 70 GPa and is subjectedto couples M = 250 kN−m at its ends. (a) What isthe resulting radius of curvature of the neutral axis? (b)Determine the maximum tensile stress due to bending.

Solution:The moment of inertia for the cross-section is:

I =bh3

12=

(0.16 m)(0.32 m)3

12= 4.369×10−4 m4

(a) Using Equation (15.10) to determine the magnitude of the radiusof curvature:

1ρ

=M

EI=

250, 000 N−m

(70×109 N/m2)(4.369×10−4 m4)= 8.174×10−3 m−1

ANS: ρ = 122.34 m

(b) The maximum normal stress due to the bending moment is:

σMAX =MyMAX

I=

(250, 000 N−m)(0.16 m)4.369×10−4 m4

ANS: σMAX = 91.6 MPa

Problem 15.2 The material of the beam in Prob-lem 15.1 will safely support a tensile stress of 180 MPaand a compressive stress of 200 MPa. Based on thesecriteria, what is the largest couple M to which the beamcan be subjected?

Solution:The symmetry of the cross-section tells us that:

(σMAX)TENSILE = (σMAX)COMPRESSIVE

The moment of inertia for the cross-section is:

I =bh3

12=

(0.16 m)(0.32 m)3

12= 4.369×10−4 m4

The maximum normal stress due to the bending moment is:

σMAX =MyMAX

I→ M =

σMAXI

yMAX=

(180×106 N/m2)(4.369×10−4 m4)0.16 m

ANS: M = 492 kN−m

Problem 15.3 The material of the beam in Prob-lem 15.1 will safely support a tensile stress of 180 MPaand a compressive stress of 200 MPa. Suppose that thebeam is rotated 90◦ about its axis, so that the width of itscross section is 0.32 m and its height is 0.16 m. What isthe largest coupleM to which the beam can be subjected?Compare your answer to the answer to Problem 15.2.

Solution:The moment of inertia for the cross-section is:

I =bh3

12=

(0.32 m)(0.16 m)3

12= 1.092×10−4 m4

Using Equation (15-12) to determine the applied moment:

σMAX =MyMAX

I→ M =

σMAXI

yMAX=

(180×106 N/m2)(1.092×10−4 m4)0.08 m

ANS: M = 246 kN−m This moment is 1/2 the result in Prob-lem 15.2.