HW 8 - Homework set 8 Solutions PDF

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Homework set 8 Solutions...

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PROBLEM 4.3

200 mm 12 mm

y

C

x

Using an allowable stress of 155 MPa, determine the largest bending moment M that can be applied to the wide-flange beam shown. Neglect the effect of fillets.

220 mm

M 8 mm 12 mm

SOLUTION Moment of inertia about x-axis:

1 (200)(12) 3  (200)(12)(104) 2 12  25.9872  10 6 mm 4

I1 

1 (8)(196)3  5.0197  106 mm4 12 I 3  I1  25.9872  106 mm4 I2 

I  I1  I2  I3  56.944  10 6 mm 4  56.944  10 6 m 4 Mc I I M  c

 

Mx 

with c 

1 (220)  110 mm  0.110 m 2

with   155  106 Pa

(56.944  106 )(155  10 6 )  80.2  103 N  m 0.110

M x  80.2 kN  m 

8 in.

PROBLEM 4.10 1 in.

1 in.

Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.

6 in. 1 in. 4 in.

A

25 kips

25 kips

B

C

20 in.

60 in.

D

20 in.

SOLUTION

A

y0



8

7.5

60



6

4

24



4

0.5



18 Yo 

A y0

2 86

86  4.778 in. 18

Neutral axis lies 4.778 in. above the base. 1 1 b h 3  A1d12  (8)(1)3  (8)(2.772)2  59.94 in4 12 1 1 12 1 1 I 2  b 2h23  A2d 22  (1)(6) 3  (6)(0.778) 2  21.63 in 4 12 12 1 1 I 3  b 3h33  A3d 23  (4)(1)3  (4)(4.278)2  73.54 in4 12 12 I  I1  I 2  I3  59.94  21.63  73.57  155.16 in4 y top  3.222 in. ybot  4.778 in. I1 

M  Pa  0 M  Pa  (25)(20)  500 kip  in.

 top  

 bot  

Mytop



(500)(3.222) 155.16



(500)( 4.778) 155.16

I Mybot I

 top  10.38 ksi (compression) 

bot  15.40 ksi (tension) 

80 mm

PROBLEM 4.19 Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

54 mm

40 mm M

SOLUTION A, mm2

y0 , mm

A y0 , mm3

d, mm

2160

27

58,320

3



1080

36

38,880

3

Σ

3240



Y 

97, 200  30 mm 3240

97,200

The neutral axis lies 30 mm above the bottom.

ytop  54  30  24 mm  0.024 m

ybot   30 mm   0.030 m

1 3 1 b1h1  A1d12  (40)(54)3  (40)(54)(3) 2  544.32  103 mm4 12 12 1 1 1 I 2  b 2h 22  A 2d 22  (40)(54) 3  (40)(54)(6) 2  213.84  103 mm4 36 36 2 I  I1  I2  758.16  103 mm4  758.16  10 9 m4 I1 

| | 

My I

|M | 

I y

Top: (tension side)

M

(120  106 )(758.16  109 )  3.7908 103 N  m 0.024

Bottom: (compression)

M

(150  106 )(758.16  109 )  3.7908 103 N  m 0.030

Choose the smaller as Mall.

M all  3.7908  10 3 N  m

M all  3.79 kN  m ...