2
CVLE 423 – Fall 2015 –Assign. #5 1 Beirut Arab University Dept. of Civil Engineering CVLE 423: Steel Design (I) FALL 2015-2016 Assignment #5 Due: November 30 -2015 Problem 1 Compute the maximum factored load that can be applied to this connection if designed as a bearing-type connection. - M20 Bolts(N-Type) - Grade 8.8 - Plates St 37 - Gas-cut edges. Problem 2 Design and detail (sketch) this double-lap splice connection for a factored load of 80 tons. Use a bolted bearing-type connection - St37- 20 mm Bolts . 0.5 cm Pu = 80 tons 30 cm Pu Pu = 80 tons PL 20 mm thick PLs 12 mm thick Pu 0.5P 0.5P P PL 12 x 260 PLs 8 x 260 P 5 cm 8 cm P 8 cm 5 cm 8 cm 8 cm 3 cm 3 cm

Assig 5

  • Upload
    khawwam

  • View
    18

  • Download
    1

Embed Size (px)

Citation preview

Page 1: Assig 5

CVLE 423 – Fall 2015 –Assign. #5

1

Beirut Arab University Dept. of Civil Engineering CVLE 423: Steel Design (I) FALL 2015-2016 Assignment #5 Due: November 30 -2015

Problem 1

Compute the maximum factored load that can be applied to this connection if designed as a bearing-type connection.

- M20 Bolts(N-Type) - Grade 8.8 - Plates St 37 - Gas-cut edges.

Problem 2

Design and detail (sketch) this double-lap splice connection for a factored load of 80 tons. Use a bolted bearing-type connection - St37- 20mm Bolts .

0.5 cm

Pu = 80tons

30 cm

Pu

Pu = 80tons

PL 20mm thick

PLs 12mm thick

Pu

0.5P

0.5P

P

PL 12 x 260

PLs 8 x 260

P

5cm

8cm P

8cm

5cm

8cm 8cm 3cm 3cm

Page 2: Assig 5

CVLE 423 – Fall 2015 –Assign. #5

2

Problem 3 For the shown truss, design and sketch the two circled connections (A

and B) as a bearing-type connections using a single row of bolts, (5/8)” – A325-X. Assume all web members are composed of single angles (4x4x5/8) – 1/2” gusset plate. Problem4

For the bearing-type connection shown in the figure below, determine how many M22 bolts (Gr8.8) are required. Assume that the bolts threads are in the plane of shear and bearing strength is controlled by the upper limit of 2.4dtFu.

2 Ls 4x4x3/8 A

10 ft

6@8 ft =48 ft

Pu = 34 k

Pu

Pu Pu

Pu B 2 Ls 5x3x1/2

St 52

Pu= 75tons

IPE 500

3

T cut from IPE 500

4