jib crane example problem

12-631 Structural Design Exam 1 S06Solution Sheet

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Jib Crane

A steel jib crane is constructed of a Beam BC pinned at Joint B and supported by a tension Member AC at Joint C as shown on the following page. Beam BC is a double channel member made up of two C9x20 channels arranged back to back to make up the beam section. Tension member AC is a single 1/2" thick by 3" wide steel bar pinned at Joint A and bolted to the Beam BC at Joint C with two 3/4" Dia. A325-N bolts. Details of the bolted connection at Joint C are shown in the Elevation and Section Views of the Detail at Joint C. The steel bar and steel channels are made of A36 steel. The dimensions and properties of the double C9x20 channels are given in the tables below.

The jib crane supports a vertical pick-up service live load, PL, which can act anywhere along

Beam BC. Find the allowable magnitude for PL considering the tension strength of Member

AC, the compression and bending strengths of Beam BC and the strength of the bolted connection at Joint C. In your analysis, the following assumptions can be made:

1. Only live loading occurs, that is, neglect the weight of the members and the dead load and impact load components of the pick-up load.

2. When the pick-up load is at Joint C, Beam BC is in "compression" only and the tension force in Member AC and compression force in BC are at their maximum values.

3. When the pick-up load is at some location between Joints B and C, the beam is in "bending" only.

4. The pinned connections at Joints A and B will not govern.

5. The compression flange of Beam BC is laterally stable.

6. Joints A, B & C are supported to prevent lateral displacement, i.e., these joints cannot move in a direction perpendicular to the Plane ABC.

7. Block shear and bending shear in the webs of the double channels will not govern.

Dimensions of each C9x20 Channel: d, in tw, in bf, in tf, in k, in 9.0 0.448 2.65 0.413 1.0

Properties of the Double C9x20 with a ½” Separation As, in

2 Zx, in3 Ix, in

4 Iy, in4 rx, in ry, in

11.74 33.8 121.8 13.0 3.22 1.05 x

ytf (avg.)bf

d

tw

k

1/2"


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A

B C

PL

1/2" x 3" Tension Bar

10' 0"

10' 0"

3"

2"

3"

1/2"

Two C9x20

Two 3/4" Dia. A325-N Bolts

DETAIL AT JOINT CElevation View Sectional View

C

kips 1000lbf


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Material Properties: Fy 36 ksi Fu 58 ksi Fub 120ksi Es 29000ksi

Length of Beam BC: Lbc 10 ft

Tension Strength of Member AC

b 3 in t1

2in sb 3 in Le 2 in db

3

4in

Tension Member Area: Ag b t Ag 1.5in2

Bolt Hole Dia.: dh db1

8in dh 0.88in

Net Area: An Ag dh t An 1.06in2

Effective Net Area: U 1 Ae U An Ae 1.06in2

Yield Limit State: Tny 0.9 Fy Ag

Tnf 0.75Fu AnFracture Limit State:

Block Shear Limit State: Agtb

2t Agt 0.75in

2

Ant Agt 0.5 dh t Ant 0.53in2

Agv sb Le t Agv 2.5in2

Anv Agv 1.5 dh t Anv 1.84in2

0.6 Fu Anv 64.16kips > Fu Ant 30.81kips

Therefore, shear fracture - tension yielding mode controls, and: Tnbs 0.75 0.6 Fu Anv Fy Agt

Strength of Bolted Connection at C

Shear Strength of Bolts:Bolt Area: Ab

4db2 Ab 0.44in

2

Bolts in double shear, therefore, mb 2

Since threads iNcluded in shear plane, the shear strength per bolt is: Rnv 0.75 0.4 Fub mb Ab Rnv 31.81kips

Tnbs 68.37kips

Tny 48.6kips

Tnf 46.22kips


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Bearing Strength of Bolts on Tension Member AC:

sb 3 in > 3 db 2.25in and Le 2in > 2.5 db 1.87in

Since the above spacing and end distances are met, then the bearing strength per bolts is: Rnb 0.75 2.4 Fu db t Rnb 39.15kips

Bolt bearing in the channel sections will not govern since the bolt load transfered to each channel is one-half of the total bolt load, and the web thickness of the channels is more that one-half the thickness of the tension member.

Rnv 31.81kips Rnb 39.15kips

Minimum Strength of the Bolted Connection: Tnb 2 min Rnv Rnb

Bolt shear controls connection strength.

Min. Strength of Tension Member and Bolts:

Tu min Tny Tnf Tnbs Tnb Fracture Limit State controls strength of AC

Equilibium of Forces at Joint CThe maximum tension in AC and compression in BC will occur when the pick-up load is at Joint C. A free-body diagram of the forces at that joint is shown in the sketch below:

Equilibrium of forces in the vertical direction gives:

Tu sin 45o - Pu = 0

Equilibrium of forces in the horizontal direction gives:

Cu - Tucos 45o = 0

Solving for Tu and Cu, we obtain:

Tu = Pu/sin 45o = 1.414 Pu

and Cu = Tu cos 45o = Pucos 45o/sin 45o = Pu

C

T

C

P

45°

u

u

u

Ultimate Pick-up Load Based on Tu: Put

Tu

1.414

Tnb 63.62kips

Tu 46.22kips

Put 32.69kips


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Compression Strength of Member BC

Required properties of the double C9x20 channels separated by 1/2"

Properties of Both Channels: Dimensions of Each Channel:

As 11.74 in2 Zx 33.8 in

3 d 9 in tw 0.448in

Ix 121.8in4 Iy 13.0 in

4 bf 2.65 in tf 0.413in

rx 3.22 in ry 1.05 in k 1 in h d 2 k

h 7in

Width to thickness ratios foreach channel:

bf

tf6.42 <

95

Fy

ksi

15.83Therefore, the sections can yield before local buckling of the flange and web.

h

tw15.63 <

253

Fy

ksi

42.17

Unbraced Length: Lbc 10 ft Kx 1.0 Ky 1.0

Since Kx = Ky = 1.0 and rx > ry, then buckling about the weak axis, y, will govern.

c < 1.5, therefore,

inelastic buckling governsc

Ky Lbc

ry

Fy

2Es

c 1.28

Fcr 0.658c2

Fy

Fcr 18.1ksi Fcr 0.85Fcr Fcr 15.39ksi

Pnc 0.85Fcr As Pnc 180.62kips

and the compression strength of the member becomes:

Cu Pnc

The maximum compression force in BC will occur when the pick-up load is at Joint C and from the above equilibrium equations, Pu = Cu.

Ultimate Pick-up Load Based on Cu: Puc Cu

Cu 180.62kips

Puc 180.62kips


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Bending Strength of BC

Check if section is compact:bf

tf6.42 <

65

Fy

ksi

10.83

h

tw15.63 <

640

Fy

ksi

106.67

Also, it is assumed that the compression flanges of the channels are laterally stable. Therefore, the channels are compact and can develop the full plastic moment

Moment Capacity of the Double C6x13 Section:

Mn 0.90Zx Fy Mn 91.26ft kips

Mu Mn Mu 91.26ft kips

B C

L

L

PuPu2

Pu2

2

Max. M = P L/4u

Pu2

As shown in the sketch above, the maximum moment in BC will occur when the pick-up load is at mid-span. In this case, the magnitude of the moment produced by the pick-up load will be: Mu =

PumLbc/4 . Solving for Pum, we obtain:

Ultimate Pick-up Load Based on Mu: Pum

4 Mu

Lbc Pum 36.5kips


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Minimum Value of Pu:

In summary, we have: Put 32.69kips Puc 180.62kips Pum 36.5kips

Therefore, Pu min Put Puc Pum Tension in AC governs.

Corrersponding Service Load:

The Live Load factor is 1.6 and theservice pick-up load becomes:

PL

Pu

1.6

Therefore, this crane should be rated for a service load of 20 kips or 10 Tons.

Pu 32.69kips

PL 20.43kips

jib crane example problem

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