A 3/4" x 10" X 5' long plate is fabricated from A36 steel and has standard holes for 3/4" bolts at each end for attachment to other structural members. The figure below shows a face view of the plate. The service loads that the member will be subject to are 140 kips of dead load and 30 kips of live load. Calculate the factored load, tensile yield strength and rupture strength. Compare the results to the factored loads and comment. Ignore block shear. +3" 3" + 3" **- - 1 1/2" 10" 5' SYM 5

A 3/4" x 10" X 5' long plate is fabricated from A36 steel and has standard holes for 3/4" bolts at each end for attachment to other structural members. The figure below shows a face view of the plate. The service loads that the member will be subject to are 140 kips of dead load and 30 kips of live load. Calculate the factored load, tensile yield strength and rupture strength. Compare the results to the factored loads and comment. Ignore block shear. +3" 3" + 3" **- - 1 1/2" 10" 5' SYM 5

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

11-1. Determine whether the W14 x 34 beam shown in Figure 11-43 is adequate for flexural rupture and flexural yielding at the cope. The steel is ASTM A992, grade 50. 1/2" W14x34 R₁ = 40kips FIGURE 11-43 Detail for Exercise 11-1
| * A's O As,web As bw Steel areas: A's-2 + 14 = 308 mm² A₁=50 mm² (area of $8) Materials : C25/30 / B420C Longitudinal reinforcements A's=2 $ 14 As,web= 2 $ 12 As= 5 0 14 bw=300 mm d=670 mm d'=30 mm k₁= 0,85 k3 = 0,85 Pb=0.0205 (from table) Transverse Reinforcements $8/200 mm (2-leg stirrups) Material coefficients Ymc:1.5 (Concrete) Yms: 1,15 (Steel) Determine the reinforcement ratio and failure type of the beam As,web= 212= 226 mm² Asw=2x50=100 mm² (Total Area of two leg stirrup) A₁= 5 + 14=770 mm²
how to solve ?? please and thank you.
Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.
A steel plate is 300 mm wide and 40 mm thick is connected using 24 mm dia bolts of 3 numbers. The net sectional area of the plate is
If you can't solve this question just skip it but don't say this is an assignment question.
Specifically state the shape and size of the structural steel action: W 36 x 182. (Show complete process or explanation in the finding the answer)
The tension member shown in the figure is connected to a gusset plate as shown with 1.210-inch diameter bolts. It is subjected to the dead and live loads as shown. Take Fy = 33.29 ksi and Fu = 52.93 ksi. Assume Ae = An. a) Draw the critical cross-section and show its dimensions using your data. b) Determine whether this member has enough strength using LRFD and the critical ASCE 7 loading combination.
Q14 What is the available tensile strength of a WT 20x74.5 made of A572 grade 50 steel? Use ASD, assume a shear lag factor (U) of 0.75, the connections are welded, and neglect block shear or other Chapter J requirements. Check both yielding and rupture. a) 800 kips b) 986 kips c) 533kips d) 656 kips
What is the working strength of a 2" bolt which is screwed up tightly in a packed joint when the allowable working stress is 12000 psi? (A) 20,120 lbs. B 20,200 lbs. C) 25,347 lbs. D) 20,443 lbs.
Show 5 failures. Show step by step solution. Steel Type: Cross section flat bar PL 12.7 x 165 with an area of 2095.5 mm2 Textbook reference Handbook of steel construction.
3-5 with clear explantion