Jib Crane Example Problem

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, in2 Zx, in3 Ix, in4 Iy, in4 rx, in ry, in 11.74 33.8 121.8 13.0 3.22 1.05 x y t f ( a v g . )b f d t w k 1 / 2 "

A

B

C

P

_L 1 / 2 " x 3 " T e n s i o n B a r

1 0 ' 0 "

1 0 ' 0 "

3 "

2 "

3 "

1 / 2 "

T w

o C

9 x 2 0

T w

o 3 / 4 " D

i a .

A

3 2 5 - N

B

o l t s

D

E T A I L A T J O

I N

T C

E

l e v a t i o n V i e w

S e c t i o n a l V i e w

C

kips:=1000lbf⋅

Material Properties: _Fy:=36 ksi⋅ _Fu:=58 ksi⋅ _Fub:=120 ksi⋅ _Es :=29000ksi⋅

Length of Beam BC: _Lbc:=10 ft⋅

Tension Strength of Member AC

b :=3 in⋅ t 1

2⋅in

:= _{sb 3 in}:= ⋅ _{Le 2 in}:= ⋅ _db 3

4⋅in

:=

Tension Member Area: _Ag:=b t⋅ _{Ag 1.5in}= 2

Bolt Hole Dia.: _{dh db} 1

8⋅in

+

:= _{dh 0.88in}=

Net Area: _{An:=Ag dh t− ⋅ An 1.06in=} 2

Effective Net Area: U:=1 _{Ae U An}:= ⋅ _{Ae 1.06in}= 2

Yield Limit State: φ_Tny:=_{0.9 Fy}⋅ ⋅_Ag

φ_Tnf :=_0.75Fu⋅ ⋅_An

Fracture Limit State:

Block Shear Limit State: _Agt b 2⋅t

:= _{Agt 0.75in}= 2

Ant:=Agt 0.5dh− ⋅ ⋅t _{Ant 0.53in}= 2

Agv :=

(

sb Le+

)

⋅t _{Agv 2.5in}= 2

Anv :=Agv 1.5dh− ⋅ ⋅t _{Anv 1.84in}= 2

0.6 Fu⋅ ⋅_Anv =64.16kips > _{Fu Ant}⋅ =30.81kips Therefore, shear fracture - tension yielding

mode controls, and: φTnbs :=0.75 0.6 Fu

(

⋅ ⋅Anv +Fy Agt⋅

)

Strength of Bolted Connection at C

Shear Strength of Bolts: _{Bolt Area:}

Ab:= π₄⋅db2 _{Ab 0.44in}= 2

Bolts in double shear, therefore, _mb :=2 Since threads iNcluded in shear plane,

the shear strength per bolt is: φRnv:=0.75 0.4 Fub⋅

(

⋅

)

⋅mb⋅Ab φRnv 31.81kips=

φ_{Tnbs 68.37kips}= φ_{Tny 48.6kips}=

Bearing Strength of Bolts on Tension Member AC:

sb 3in= > _{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: φRnb:=0.75 2.4 Fu⋅

(

⋅

)

⋅db⋅t φRnb 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.

φ_{Rnv 31.81kips}= φ_{Rnb 39.15kips}=

Minimum Strength of the Bolted Connection: φ_Tnb:=2 min⋅

(

φ_Rnv,φ_Rnb

)

Bolt shear controls connection strength. Min. Strength of Tension Member and Bolts:

Tu:=min

(

φ_Tny,φ_Tnf,φ_Tnbs,φ_Tnb

)

Fracture Limit State controls strength of AC

Equilibium of Forces at Joint C

The 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

4 5 °

u

u

u

Ultimate Pick-up Load Based on Tu: _Put Tu

1.414

:=

φ_{Tnb 63.62kips}=

Tu 46.22kips=

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:=11.74in⋅ 2 _{Zx 33.8in}:= ⋅ 3 d:=9 in⋅ _{tw 0.448in}:= ⋅

Ix 121.8in:= ⋅ 4 _Iy:=13.0in⋅ 4 _bf:=2.65in⋅ _{tf 0.413in}:= ⋅

rx 3.22in:= ⋅ _ry :=1.05in⋅ k:=1 in⋅ h :=d −2 k⋅

h =7 in Width to thickness ratios for

each channel: bf tf =6.42 < 95 Fy ksi 15.83 =

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

h tw =15.63 < 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.

λ _{c < 1.5, therefore,}

inelastic buckling governs

λ_c Ky Lbc⋅ ry Fy π2⋅_Es ⋅ := λ_c=1.28 Fcr 0.658λc 2 Fy ⋅









:= _{Fcr 18.1ksi}= φ_Fcr :=_0.85Fcr⋅ φ_{Fcr 15.39ksi}= φ_Pnc:=_0.85Fcr⋅ ⋅_As φ_{Pnc 180.62kips}=

and the compression strength of the member becomes: Cu :=φ_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=

Bending Strength of BC

Check if section is compact: bf

tf =6.42 < 65 Fy ksi 10.83 = h tw =15.63 < 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:

φ_Mn:=_0.90Zx⋅ ⋅_Fy φ_{Mn 91.26ft kips}= ⋅ Mu:=φ_{Mn Mu 91.26ft kips}= ⋅

B

_C

L

L

P

_u

P

_u 2

P

_u 2 2 M a x . M_u = P L / 4

P

_u 2

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

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 the

service 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=